Koshe Dekhi 9.3 Class 10 | দ্বিঘাত করণী কষে দেখি 9.3 | গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3 | WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Dighat Koroni (Surds) Solution in Bengali
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Koshe Dekhi 9.3 Class 10|দ্বিঘাত করণী কষে দেখি 9.3|গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3|WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Surds Solution in Bengali
কষে দেখি - 9.3
Koshe Dekhi 9.3 Class 10|দ্বিঘাত করণী কষে দেখি 9.3|গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3|WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Surds Solution in Bengali
1.
(a) \(m + \frac{1}{m} = \sqrt 3 \) হলে \({\rm{ (i) }}{m^2} + \frac{1}{{{m^2}}}\) এবং \({\rm{ (ii) }}{m^3} + \frac{1}{{{m^3}}}\) এদের সরলতম মান নির্ণয় করি ।
(i) \(m + \frac{1}{m} = \sqrt 3 \)
দুদিকের বর্ণ করলে, \({\left( {m + \frac{1}{m}} \right)^2} = {(\sqrt 3 )^2}\)
বা, \({m^2} + 2 \cdot m \cdot \frac{1}{m} + \frac{1}{{{m^2}}} = 3\)
বা, \({m^2} + 2 + \frac{1}{{{m^2}}} = 3\)
\(\therefore {m^2} + \frac{1}{{{m^2}}} = 3 - 2 = 1\)
\(\therefore\) নির্ণেয় মান \(= 1\) (ii) \({m^3} + \frac{1}{{{m^3}}} = {\left( {m + \frac{1}{m}} \right)^3} - 3m\frac{1}{m}\left( {m + \frac{1}{m}} \right)\)
\(=(\sqrt{3})^{3}-3 \sqrt{3}\)
\(=3 \sqrt{3}-3 \sqrt{3}\)
\(=0\) \(\therefore\) নির্ণেয় মান \(= 0\)
দুদিকের বর্ণ করলে, \({\left( {m + \frac{1}{m}} \right)^2} = {(\sqrt 3 )^2}\)
বা, \({m^2} + 2 \cdot m \cdot \frac{1}{m} + \frac{1}{{{m^2}}} = 3\)
বা, \({m^2} + 2 + \frac{1}{{{m^2}}} = 3\)
\(\therefore {m^2} + \frac{1}{{{m^2}}} = 3 - 2 = 1\)
\(\therefore\) নির্ণেয় মান \(= 1\) (ii) \({m^3} + \frac{1}{{{m^3}}} = {\left( {m + \frac{1}{m}} \right)^3} - 3m\frac{1}{m}\left( {m + \frac{1}{m}} \right)\)
\(=(\sqrt{3})^{3}-3 \sqrt{3}\)
\(=3 \sqrt{3}-3 \sqrt{3}\)
\(=0\) \(\therefore\) নির্ণেয় মান \(= 0\)
(b) দেখাই যে, \(\frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} - \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} = 2\sqrt {15} \)
\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\frac{(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}\)
\(=\frac{(\sqrt{5})^{2}+2 \cdot \sqrt{5} \cdot \sqrt{3}+(\sqrt{3})^{2}-\left\{(\sqrt{5})^{2}-2 \cdot \sqrt{5} \cdot \sqrt{3}+(\sqrt{3})^{2}\right\}}{5-3}\) \(=\frac{5+3+2 \sqrt{15}-5-3+2 \sqrt{15}}{5-3}\)
\(=\frac{4 \sqrt{15}}{2}=2 \sqrt{15}\)
\(\therefore\) নির্ণেয় সরলতম মান \(=2 \sqrt{15}\)
\(=\frac{(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}\)
\(=\frac{(\sqrt{5})^{2}+2 \cdot \sqrt{5} \cdot \sqrt{3}+(\sqrt{3})^{2}-\left\{(\sqrt{5})^{2}-2 \cdot \sqrt{5} \cdot \sqrt{3}+(\sqrt{3})^{2}\right\}}{5-3}\) \(=\frac{5+3+2 \sqrt{15}-5-3+2 \sqrt{15}}{5-3}\)
\(=\frac{4 \sqrt{15}}{2}=2 \sqrt{15}\)
\(\therefore\) নির্ণেয় সরলতম মান \(=2 \sqrt{15}\)
2.
(a) সরল করি \(\frac{{\sqrt 2 (2 + \sqrt 3 )}}{{\sqrt 3 (\sqrt 3 + 1)}} - \frac{{\sqrt 2 (2 - \sqrt 3 )}}{{\sqrt 3 (\sqrt 3 - 1)}}\)
\(\frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{3}(\sqrt{3}+1)}-\frac{\sqrt{2}(2-\sqrt{3})}{\sqrt{3}(\sqrt{3}-1)}\)
\(=\frac{\sqrt{2}(2+\sqrt{3})(\sqrt{3}-1)}{\sqrt{3}(\sqrt{3}+1)(\sqrt{3}-1)}-\frac{\sqrt{2}(2-\sqrt{3})(\sqrt{3}+1)}{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}\)
\(=\frac{\sqrt{2}(2 \sqrt{3}-2+3-\sqrt{3})}{\sqrt{3}\left\{(\sqrt{3})^{2}-(1)^{2}\right\}}-\frac{\sqrt{2}(2 \sqrt{3}+2-3-\sqrt{3})}{\sqrt{3}\left\{(\sqrt{3})^{2}-(1)^{2}\right\}}\)
\(=\frac{\sqrt{2}(\sqrt{3}+1)}{\sqrt{3}(3-1)}-\frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{3}(3-1)}=\frac{\sqrt{2}(\sqrt{3}+1)}{2 \sqrt{3}}-\frac{\sqrt{2}(\sqrt{3}-1)}{2 \sqrt{3}}\)
\(=\frac{\sqrt{2}(\sqrt{3}+1-\sqrt{3}+1)}{2 \sqrt{3}}=\frac{2 \sqrt{2}}{2 \sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{\sqrt{6}}{3}\)
\(\therefore\) নির্ণেয় মান \(=\frac{\sqrt{6}}{3}\)
\(=\frac{\sqrt{2}(2+\sqrt{3})(\sqrt{3}-1)}{\sqrt{3}(\sqrt{3}+1)(\sqrt{3}-1)}-\frac{\sqrt{2}(2-\sqrt{3})(\sqrt{3}+1)}{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}\)
\(=\frac{\sqrt{2}(2 \sqrt{3}-2+3-\sqrt{3})}{\sqrt{3}\left\{(\sqrt{3})^{2}-(1)^{2}\right\}}-\frac{\sqrt{2}(2 \sqrt{3}+2-3-\sqrt{3})}{\sqrt{3}\left\{(\sqrt{3})^{2}-(1)^{2}\right\}}\)
\(=\frac{\sqrt{2}(\sqrt{3}+1)}{\sqrt{3}(3-1)}-\frac{\sqrt{2}(\sqrt{3}-1)}{\sqrt{3}(3-1)}=\frac{\sqrt{2}(\sqrt{3}+1)}{2 \sqrt{3}}-\frac{\sqrt{2}(\sqrt{3}-1)}{2 \sqrt{3}}\)
\(=\frac{\sqrt{2}(\sqrt{3}+1-\sqrt{3}+1)}{2 \sqrt{3}}=\frac{2 \sqrt{2}}{2 \sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{\sqrt{6}}{3}\)
\(\therefore\) নির্ণেয় মান \(=\frac{\sqrt{6}}{3}\)
(b) সরল করি : \(\frac{{3\sqrt 7 }}{{\sqrt 5 + \sqrt 2 }} - \frac{{5\sqrt 5 }}{{\sqrt 2 + \sqrt 7 }} + \frac{{2\sqrt 2 }}{{\sqrt 7 + \sqrt 5 }}\)
প্রদত্ত রাশিমালা
\(=\frac{3 \sqrt{7}}{\sqrt{2}+\sqrt{5}}-\frac{5 \sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2 \sqrt{2}}{\sqrt{5}+\sqrt{7}}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{5-2}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{7-2}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{7-5}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{3}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{5}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{2}\)
\(=\sqrt{7}(\sqrt{5}-\sqrt{2})-\sqrt{5}(\sqrt{7}-\sqrt{2})+\sqrt{2}(\sqrt{7}-\sqrt{5})\)
\(=\sqrt{35}-\sqrt{14}-\sqrt{35}+\sqrt{10}+\sqrt{14}-\sqrt{10}=0\)
\(\therefore\) নির্ণেয় মান \(=0\)
\(=\frac{3 \sqrt{7}}{\sqrt{2}+\sqrt{5}}-\frac{5 \sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2 \sqrt{2}}{\sqrt{5}+\sqrt{7}}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^{2}-(\sqrt{2})^{2}}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{(\sqrt{7})^{2}-(\sqrt{2})^{2}}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{5-2}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{7-2}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{7-5}\)
\(=\frac{3 \sqrt{7}(\sqrt{5}-\sqrt{2})}{3}-\frac{5 \sqrt{5}(\sqrt{7}-\sqrt{2})}{5}+\frac{2 \sqrt{2}(\sqrt{7}-\sqrt{5})}{2}\)
\(=\sqrt{7}(\sqrt{5}-\sqrt{2})-\sqrt{5}(\sqrt{7}-\sqrt{2})+\sqrt{2}(\sqrt{7}-\sqrt{5})\)
\(=\sqrt{35}-\sqrt{14}-\sqrt{35}+\sqrt{10}+\sqrt{14}-\sqrt{10}=0\)
\(\therefore\) নির্ণেয় মান \(=0\)
সরল করি\({\rm{(c) }}\frac{{4\sqrt 3 }}{{2 - \sqrt 2 }} - \frac{{30}}{{4\sqrt 3 - \sqrt {18} }} - \frac{{\sqrt {18} }}{{3 - \sqrt {12} }}\)
প্রদত্ত রাশিমালা
\(\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-2 \sqrt{3}}\)
\(=\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-3 \sqrt{2}}-\frac{3 \sqrt{2}}{3-2 \sqrt{3}}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{(4 \sqrt{3}-3 \sqrt{2})(4 \sqrt{3}+3 \sqrt{2})}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{(3-2 \sqrt{3})(3+2 \sqrt{3})}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{(4 \sqrt{3})^{2}-(3 \sqrt{2})^{2}}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{(3)^{2}-(2 \sqrt{3})^{2}}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{4-2}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{48-18}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{9-12}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{2}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{30}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{-3}\)
\(=2 \sqrt{3}(2+\sqrt{2})-(4 \sqrt{3}+3 \sqrt{2})+\sqrt{2}(3+2 \sqrt{3})\)
\(=4 \sqrt{3}+2 \sqrt{6}-4 \sqrt{3}-3 \sqrt{2}+3 \sqrt{2}+2 \sqrt{6}\)
\(=4 \sqrt{6}\), এটিই নির্ণেয় মান। \(\therefore\) নির্ণেয় মান \(=4 \sqrt{6}\)
\(\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-2 \sqrt{3}}\)
\(=\frac{4 \sqrt{3}}{2-\sqrt{2}}-\frac{30}{4 \sqrt{3}-3 \sqrt{2}}-\frac{3 \sqrt{2}}{3-2 \sqrt{3}}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{(4 \sqrt{3}-3 \sqrt{2})(4 \sqrt{3}+3 \sqrt{2})}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{(3-2 \sqrt{3})(3+2 \sqrt{3})}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{(2)^{2}-(\sqrt{2})^{2}}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{(4 \sqrt{3})^{2}-(3 \sqrt{2})^{2}}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{(3)^{2}-(2 \sqrt{3})^{2}}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{4-2}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{48-18}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{9-12}\)
\(=\frac{4 \sqrt{3}(2+\sqrt{2})}{2}-\frac{30(4 \sqrt{3}+3 \sqrt{2})}{30}-\frac{3 \sqrt{2}(3+2 \sqrt{3})}{-3}\)
\(=2 \sqrt{3}(2+\sqrt{2})-(4 \sqrt{3}+3 \sqrt{2})+\sqrt{2}(3+2 \sqrt{3})\)
\(=4 \sqrt{3}+2 \sqrt{6}-4 \sqrt{3}-3 \sqrt{2}+3 \sqrt{2}+2 \sqrt{6}\)
\(=4 \sqrt{6}\), এটিই নির্ণেয় মান। \(\therefore\) নির্ণেয় মান \(=4 \sqrt{6}\)
সরল করি\({\rm{(d) }}\frac{{3\sqrt 2 }}{{\sqrt 3 + \sqrt 6 }} - \frac{{4\sqrt 3 }}{{\sqrt 6 + \sqrt 2 }} + \frac{{\sqrt 6 }}{{\sqrt 2 + \sqrt 3 }}\)
\(\frac{3 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3})^{2}-(\sqrt{6})^{2}}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^{2}-(\sqrt{2})^{2}}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{3-6}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{2-3}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}\)
\(=-\sqrt{2}(\sqrt{3}-\sqrt{6})-\sqrt{3}(\sqrt{6}-\sqrt{2})-\sqrt{6}(\sqrt{2}-\sqrt{3})\)
\(=-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}=0 \)
\(\therefore\) নির্ণেয় মান \(=0\)।
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{(\sqrt{3})^{2}-(\sqrt{6})^{2}}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{(\sqrt{6})^{2}-(\sqrt{2})^{2}}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{3-6}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{2-3}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}\)
\(=-\sqrt{2}(\sqrt{3}-\sqrt{6})-\sqrt{3}(\sqrt{6}-\sqrt{2})-\sqrt{6}(\sqrt{2}-\sqrt{3})\)
\(=-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}=0 \)
\(\therefore\) নির্ণেয় মান \(=0\)।
3. যদি \({\rm{x = 2 , y = 3}}\) এবং \(z = 6\) হয় তবে, \(\frac{{3\sqrt x }}{{\sqrt y + \sqrt z }} - \frac{{4\sqrt y }}{{\sqrt z + \sqrt x }} + \frac{{\sqrt z }}{{\sqrt x + \sqrt y }}\) এর মান হিসাব করে লিখি।
=\(\frac{3 \sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4 \sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{3 \sqrt{x}(\sqrt{y}-\sqrt{z})}{(\sqrt{y}+\sqrt{z})(\sqrt{y}-\sqrt{z})}-\frac{4 \sqrt{y}(\sqrt{z}-\sqrt{x})}{(\sqrt{z}+\sqrt{x})(\sqrt{z}-\sqrt{x})}+\frac{\sqrt{z}(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}\)
\(=\frac{3 \sqrt{x}(\sqrt{y}-\sqrt{z})}{(\sqrt{y})^{2}-(\sqrt{z})^{2}}-\frac{4 \sqrt{y}(\sqrt{z}-\sqrt{x})}{(\sqrt{z})^{2}-(\sqrt{x})^{2}}+\frac{\sqrt{z}(\sqrt{x}-\sqrt{y})}{(\sqrt{x})^{2}-(\sqrt{y})^{2}}\)
\(=\frac{3 \sqrt{x}(\sqrt{y}-\sqrt{z})}{y-z}-\frac{4 \sqrt{y}(\sqrt{z}-\sqrt{x})}{z-x}+\frac{\sqrt{z}(\sqrt{x}-\sqrt{y})}{x-y}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{3-6}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{2-3}\) [\(x, y\) এবং \(z\)-এর মান বসিয়ে]
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}\)
\(=-\sqrt{2}(\sqrt{3}-\sqrt{6})-\sqrt{3}(\sqrt{6}-\sqrt{2})-\sqrt{6}(\sqrt{2}-\sqrt{3})\)
\(=-\sqrt{6}+\sqrt{12} \cdot \sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}=0 \)
\(\therefore\) নির্ণেয় মান \(=0\) ।
\(=\frac{3 \sqrt{x}(\sqrt{y}-\sqrt{z})}{(\sqrt{y}+\sqrt{z})(\sqrt{y}-\sqrt{z})}-\frac{4 \sqrt{y}(\sqrt{z}-\sqrt{x})}{(\sqrt{z}+\sqrt{x})(\sqrt{z}-\sqrt{x})}+\frac{\sqrt{z}(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}\)
\(=\frac{3 \sqrt{x}(\sqrt{y}-\sqrt{z})}{(\sqrt{y})^{2}-(\sqrt{z})^{2}}-\frac{4 \sqrt{y}(\sqrt{z}-\sqrt{x})}{(\sqrt{z})^{2}-(\sqrt{x})^{2}}+\frac{\sqrt{z}(\sqrt{x}-\sqrt{y})}{(\sqrt{x})^{2}-(\sqrt{y})^{2}}\)
\(=\frac{3 \sqrt{x}(\sqrt{y}-\sqrt{z})}{y-z}-\frac{4 \sqrt{y}(\sqrt{z}-\sqrt{x})}{z-x}+\frac{\sqrt{z}(\sqrt{x}-\sqrt{y})}{x-y}\)
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{3-6}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{6-2}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{2-3}\) [\(x, y\) এবং \(z\)-এর মান বসিয়ে]
\(=\frac{3 \sqrt{2}(\sqrt{3}-\sqrt{6})}{-3}-\frac{4 \sqrt{3}(\sqrt{6}-\sqrt{2})}{4}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{-1}\)
\(=-\sqrt{2}(\sqrt{3}-\sqrt{6})-\sqrt{3}(\sqrt{6}-\sqrt{2})-\sqrt{6}(\sqrt{2}-\sqrt{3})\)
\(=-\sqrt{6}+\sqrt{12} \cdot \sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}=0 \)
\(\therefore\) নির্ণেয় মান \(=0\) ।
4. \(x = \sqrt 7 + \sqrt 6 \) হলে \({\rm{ (i) }}x - \frac{1}{x}\) \({\rm{ (ii) }}x + \frac{1}{x}\) \({\rm{ (iii) }}{x^2} + \frac{1}{{{x^2}}}\) \({\rm{ (iv) }}{x^3} + \frac{1}{{{x^3}}}\) এদের সরলতম মান নির্ণয় করি।
\(x = \sqrt 7 + \sqrt 6 \)
\(\therefore \frac{1}{x}\)
\(= \frac{1}{{\sqrt 7 + \sqrt 6 }}\)
\(=\frac{1(\sqrt{7}-\sqrt{6})}{(\sqrt{7}+\sqrt{6})(\sqrt{7}-\sqrt{6})}\)
\(=\frac{\sqrt{7}-\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}\)
\(= \frac{{\sqrt 7 - \sqrt 6 }}{{7 - 6}}\)
\(= \frac{{\sqrt 7 - \sqrt 6 }}{1}\)
\( = \sqrt 7 - \sqrt 6 \)
(i) \(x - \frac{1}{x} = (\sqrt 7 + \sqrt 6 ) - (\sqrt 7 - \sqrt 6 )\)
\(= \sqrt 7 + \sqrt 6 - \sqrt 7 + \sqrt 6 \)
\( = 2\sqrt 6 \)
\(\therefore\) নির্ণেয় সরলতম মান \( = 2\sqrt 6 \)
(ii) \(x + \frac{1}{x}\)
\( = \sqrt 7 + \sqrt 6 + \sqrt 7 - \sqrt 6\)
\( = 2\sqrt 7 \)
\(\therefore\) নির্ণেয় সরলতম মান \( = 2\sqrt 7 \)
(iii) \({x^2} + \frac{1}{{{x^2}}}\)
\(=(x)^{2}+\left(\frac{1}{x}\right)^{2}\)
\(= {\left( {x + \frac{1}{{{x}}}} \right)^2} - 2 \cdot x \cdot \frac{1}{x}\)
\(= {(2\sqrt 7 )^2} – 2\) \(\left[\because x+\frac{1}{x}=2 \sqrt{7}\right]\)
\( = 28 – 2\)
\( = 26\)
\(\therefore\) নির্ণেয় সরলতম মান \( = 26\)
(iv) \({x^3} + \frac{1}{{{x^3}}}\)
\(= {\left( {x + \frac{1}{x}} \right)^3} - 3 \cdot x \cdot \frac{1}{x}\left( {x + \frac{1}{x}}\right)\)
\(= {(2\sqrt 7 )^3} - 3 \cdot 2\sqrt 7 = 56\sqrt 7 - 6\sqrt 7\)
\(= 50\sqrt 7 \)
\(\therefore\) নির্ণেয় সরলতম মান \(= 50\sqrt 7 \)
\(\therefore \frac{1}{x}\)
\(= \frac{1}{{\sqrt 7 + \sqrt 6 }}\)
\(=\frac{1(\sqrt{7}-\sqrt{6})}{(\sqrt{7}+\sqrt{6})(\sqrt{7}-\sqrt{6})}\)
\(=\frac{\sqrt{7}-\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}\)
\(= \frac{{\sqrt 7 - \sqrt 6 }}{{7 - 6}}\)
\(= \frac{{\sqrt 7 - \sqrt 6 }}{1}\)
\( = \sqrt 7 - \sqrt 6 \)
(i) \(x - \frac{1}{x} = (\sqrt 7 + \sqrt 6 ) - (\sqrt 7 - \sqrt 6 )\)
\(= \sqrt 7 + \sqrt 6 - \sqrt 7 + \sqrt 6 \)
\( = 2\sqrt 6 \)
\(\therefore\) নির্ণেয় সরলতম মান \( = 2\sqrt 6 \)
(ii) \(x + \frac{1}{x}\)
\( = \sqrt 7 + \sqrt 6 + \sqrt 7 - \sqrt 6\)
\( = 2\sqrt 7 \)
\(\therefore\) নির্ণেয় সরলতম মান \( = 2\sqrt 7 \)
(iii) \({x^2} + \frac{1}{{{x^2}}}\)
\(=(x)^{2}+\left(\frac{1}{x}\right)^{2}\)
\(= {\left( {x + \frac{1}{{{x}}}} \right)^2} - 2 \cdot x \cdot \frac{1}{x}\)
\(= {(2\sqrt 7 )^2} – 2\) \(\left[\because x+\frac{1}{x}=2 \sqrt{7}\right]\)
\( = 28 – 2\)
\( = 26\)
\(\therefore\) নির্ণেয় সরলতম মান \( = 26\)
(iv) \({x^3} + \frac{1}{{{x^3}}}\)
\(= {\left( {x + \frac{1}{x}} \right)^3} - 3 \cdot x \cdot \frac{1}{x}\left( {x + \frac{1}{x}}\right)\)
\(= {(2\sqrt 7 )^3} - 3 \cdot 2\sqrt 7 = 56\sqrt 7 - 6\sqrt 7\)
\(= 50\sqrt 7 \)
\(\therefore\) নির্ণেয় সরলতম মান \(= 50\sqrt 7 \)
Koshe Dekhi 9.3 Class 10|দ্বিঘাত করণী কষে দেখি 9.3|গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3|WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Surds Solution in Bengali
5. সরল করি \(\frac{{x + \sqrt {{x^2} - 1} }}{{x - \sqrt {{x^2} - 1} }} + \frac{{x - \sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }}\) সরলফল 14 হলে, \(x\)-এর মান কী কী হবে হিসাব করে লিখি।
এখানে,
\(\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}=14\)
বা, \(\frac{\left(x+\sqrt{x^{2}-1}\right)^{2}+\left(x+\sqrt{x^{2}-1}\right)^{2}}{\left(x-\sqrt{x^{2}-1}\right)\left(x+\sqrt{x^{2}-1}\right)}=14\)
বা, \(\frac{\left(x+\sqrt{x^{2}-1}\right)^{2}+\left(x-\sqrt{x^{2}-1}\right)^{2}}{x^{2}-\left(\sqrt{x^{2}-1}\right)^{2}}=14\)
বা, \(\frac{x^{2}+x^{2}-1+2 x \sqrt{x^{2}-1}+x^{2}+x^{2}-1-2 x \sqrt{x^{2}-1}}{x^{2}-x^{2}+1}=14\)
বা, \(\frac{4 x^{2}-2}{1}=14\)
বা, \(4 x^{2}-2=14\)
বা, \(4 x^{2}=14+2\)
বা, \(4 x^{2}=16\)
বা, \(x^{2}=\frac{16}{4}\)
বা, \(x^{2}=4\)
বা, \(x=\pm \sqrt{4}\)
বা, \(x=\pm 2\)
\(\therefore x\) এর নির্ণেয় মান \(=\pm 2\)
\(\frac{x+\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}+\frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}}=14\)
বা, \(\frac{\left(x+\sqrt{x^{2}-1}\right)^{2}+\left(x+\sqrt{x^{2}-1}\right)^{2}}{\left(x-\sqrt{x^{2}-1}\right)\left(x+\sqrt{x^{2}-1}\right)}=14\)
বা, \(\frac{\left(x+\sqrt{x^{2}-1}\right)^{2}+\left(x-\sqrt{x^{2}-1}\right)^{2}}{x^{2}-\left(\sqrt{x^{2}-1}\right)^{2}}=14\)
বা, \(\frac{x^{2}+x^{2}-1+2 x \sqrt{x^{2}-1}+x^{2}+x^{2}-1-2 x \sqrt{x^{2}-1}}{x^{2}-x^{2}+1}=14\)
বা, \(\frac{4 x^{2}-2}{1}=14\)
বা, \(4 x^{2}-2=14\)
বা, \(4 x^{2}=14+2\)
বা, \(4 x^{2}=16\)
বা, \(x^{2}=\frac{16}{4}\)
বা, \(x^{2}=4\)
বা, \(x=\pm \sqrt{4}\)
বা, \(x=\pm 2\)
\(\therefore x\) এর নির্ণেয় মান \(=\pm 2\)
6. যদি \(a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}\) ও \(b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\) হয়, তবে নীচের মানগুলি নির্ণয় করি।\({\rm{(i) }}\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}\) \({\rm{ (ii) }}\frac{{{{(a - b)}^3}}}{{{{(a + b)}^3}}}{\rm{ }}\) \({\rm{ (iii) }}\frac{{3{a^2} + 5ab + 3{b^2}}}{{3{a^2} - 5ab + 3{b^2}}}\) \({\rm{ (iv) }}\frac{{{a^3} + {b^3}}}{{{a^3} - {b^3}}}\)
প্রথমে \(a\) এবং \(b\)-এর মান নির্ণয় করে পাওয়া যায়।
\(a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\)
\(= \frac{{{{(\sqrt 5 + 1)}^2} + {{(\sqrt 5 - 1)}^2}}}{{(\sqrt 5 - 1)(\sqrt 5 + 1)}}\)
\( = \frac{{5 + 2\sqrt 5 + 1 + (5 - 2\sqrt 5 + 1)}}{{{{(\sqrt 5 )}^2} - {{(1)}^2}}}\)
\(= \frac{{12}}{{5 - 1}}\)
\(= \frac{{12}}{4}\)
\(= 3\)
এবং \(a - b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} - \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\)
\(=\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}+1)}\)
\(= \frac{{{{(\sqrt 5 + 1)}^2} - {{(\sqrt 5 - 1)}^2}}}{{{{(\sqrt 5 )}^2} - {{(1)}^2}}}\)
\(=\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{5-1}\)
\( = \frac{{5 + 1 + 2\sqrt 5 - (5 - 2\sqrt 5 + 1)}}{4}\)
\(=\frac{5+1+2 \sqrt{5}-5+2 \sqrt{5}-1}{4}\)
\(= \frac{{4\sqrt 5 }}{4}\)
\( = \sqrt 5 \)
আবার, \(ab = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\)
\(= 1\)
(i) \(\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}\)
\(= \frac{{{a^2} + 2ab + {b^2} - ab}}{{{a^2} + 2ab + {b^2} - 3ab}}\)
\(= \frac{{{{(a + b)}^2} - ab}}{{{{(a + b)}^2} - 3ab}}\)
\( = \frac{{{{(3)}^2} - 1}}{{{{(3)}^2} - 3 \times 1}}\) \([\because a+b=3]\)
\(= \frac{{{{(3)}^2} - 1}}{{{{(3)}^2} - 3}}\)
\(= \frac{{9 - 1}}{{9 - 3}}\)
\(= \frac{8}{6}\)
\(= \frac{4}{3}\)
\(= 1\frac{1}{3}\) (উত্তর)
\(\therefore\) নির্ণেয় মান \(= 1\frac{1}{3}\)
(ii) \(\frac{{{{(a - b)}^3}}}{{{{(a + b)}^3}}}\)
\(= \frac{{{{(\sqrt 5 )}^3}}}{{{{(3)}^3}}}\)\([\because a-b=\sqrt{5}, a+b=3]\)
\(= \frac{{5\sqrt 5 }}{{27}}\) (উত্তর)
\(\therefore\) নির্ণেয় মান \(= \frac{{5\sqrt 5 }}{{27}}\)
(iii) \(\frac{{3{a^2} + 5ab + 3{b^2}}}{{3{a^2} - 5ab + 3{b^2}}}\)
\(\therefore \frac{{3{a^2} + 5ab + 3{b^2}}}{{3{a^2} - 5ab + 3{b^2}}}\)
\( = \frac{{3\left( {{a^2} + {b^2}} \right) + 5ab}}{{3\left( {{a^2} + {b^2}} \right) - 5ab}}\)
= \(\frac{{3\left\{ {{{(a + b)}^2} - 2ab} \right\} + 5ab}}{{3\left\{ {{{(a + b)}^2} - 2ab} \right\} - 5ab}}\)
\(= \frac{{3\left\{ {{{(3)}^2} - 2 \times 1} \right\} + 5 \times 1}}{{3\left\{ {{{(3)}^2} - 2 \times 1} \right\} - 5 \times 1}}\) \(\left[\begin{array}{c}\because(a+b)=3] \\ a b=1\end{array}\right.\)
\( = \frac{{3\{ 9 - 2\} + 5}}{{3\{ 9 - 2\} - 5}}\)
\(= \frac{{3 \times 7 + 5}}{{3 \times 7 - 5}}\)
\(= \frac{{21 + 5}}{{21 - 5}}\)
\(= \frac{{26}}{{16}}\)
\(= \frac{{13}}{8}\)
\(= 1\frac{5}{8}\) (উত্তর)
\(\therefore\) নির্ণেয় মান \(= 1\frac{5}{8}\)
(iv) \(\frac{{{a^3} + {b^3}}}{{{a^3} - {b^3}}}\)
সমাধান \(\frac{{{a^3} + {b^3}}}{{{a^3} - {b^3}}}\)
\(= \frac{{{{(a + b)}^3} - 3ab(a + b)}}{{{{(a - b)}^3} + 3ab(a - b)}}\)
\( = \frac{{{{(3)}^3} - 3 \times 1 \times 3}}{{{{(\sqrt 5 )}^3} + 3 \times 1 \times \sqrt 5 }}\)\([\because a+b=3, a b=1]\)
\(= \frac{{27 - 9}}{{5\sqrt 5 + 3\sqrt 5 }}\)
\(= \frac{{18}}{{8\sqrt 5 }}\)
\(= \frac{9}{{4\sqrt 5 }}\)
\(= \frac{{9 \times \sqrt 5 }}{{4\sqrt 5 \times \sqrt 5 }}\)
\(= \frac{{9\sqrt 5 }}{{20}}\) \(\therefore\) নির্ণেয় মান \(= \frac{{9\sqrt 5 }}{{20}}\)
\(a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\)
\(= \frac{{{{(\sqrt 5 + 1)}^2} + {{(\sqrt 5 - 1)}^2}}}{{(\sqrt 5 - 1)(\sqrt 5 + 1)}}\)
\( = \frac{{5 + 2\sqrt 5 + 1 + (5 - 2\sqrt 5 + 1)}}{{{{(\sqrt 5 )}^2} - {{(1)}^2}}}\)
\(= \frac{{12}}{{5 - 1}}\)
\(= \frac{{12}}{4}\)
\(= 3\)
এবং \(a - b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} - \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\)
\(=\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{(\sqrt{5}-1)(\sqrt{5}+1)}\)
\(= \frac{{{{(\sqrt 5 + 1)}^2} - {{(\sqrt 5 - 1)}^2}}}{{{{(\sqrt 5 )}^2} - {{(1)}^2}}}\)
\(=\frac{(\sqrt{5}+1)^{2}-(\sqrt{5}-1)^{2}}{5-1}\)
\( = \frac{{5 + 1 + 2\sqrt 5 - (5 - 2\sqrt 5 + 1)}}{4}\)
\(=\frac{5+1+2 \sqrt{5}-5+2 \sqrt{5}-1}{4}\)
\(= \frac{{4\sqrt 5 }}{4}\)
\( = \sqrt 5 \)
আবার, \(ab = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}\)
\(= 1\)
(i) \(\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}\)
\(= \frac{{{a^2} + 2ab + {b^2} - ab}}{{{a^2} + 2ab + {b^2} - 3ab}}\)
\(= \frac{{{{(a + b)}^2} - ab}}{{{{(a + b)}^2} - 3ab}}\)
\( = \frac{{{{(3)}^2} - 1}}{{{{(3)}^2} - 3 \times 1}}\) \([\because a+b=3]\)
\(= \frac{{{{(3)}^2} - 1}}{{{{(3)}^2} - 3}}\)
\(= \frac{{9 - 1}}{{9 - 3}}\)
\(= \frac{8}{6}\)
\(= \frac{4}{3}\)
\(= 1\frac{1}{3}\) (উত্তর)
\(\therefore\) নির্ণেয় মান \(= 1\frac{1}{3}\)
(ii) \(\frac{{{{(a - b)}^3}}}{{{{(a + b)}^3}}}\)
\(= \frac{{{{(\sqrt 5 )}^3}}}{{{{(3)}^3}}}\)\([\because a-b=\sqrt{5}, a+b=3]\)
\(= \frac{{5\sqrt 5 }}{{27}}\) (উত্তর)
\(\therefore\) নির্ণেয় মান \(= \frac{{5\sqrt 5 }}{{27}}\)
(iii) \(\frac{{3{a^2} + 5ab + 3{b^2}}}{{3{a^2} - 5ab + 3{b^2}}}\)
\(\therefore \frac{{3{a^2} + 5ab + 3{b^2}}}{{3{a^2} - 5ab + 3{b^2}}}\)
\( = \frac{{3\left( {{a^2} + {b^2}} \right) + 5ab}}{{3\left( {{a^2} + {b^2}} \right) - 5ab}}\)
= \(\frac{{3\left\{ {{{(a + b)}^2} - 2ab} \right\} + 5ab}}{{3\left\{ {{{(a + b)}^2} - 2ab} \right\} - 5ab}}\)
\(= \frac{{3\left\{ {{{(3)}^2} - 2 \times 1} \right\} + 5 \times 1}}{{3\left\{ {{{(3)}^2} - 2 \times 1} \right\} - 5 \times 1}}\) \(\left[\begin{array}{c}\because(a+b)=3] \\ a b=1\end{array}\right.\)
\( = \frac{{3\{ 9 - 2\} + 5}}{{3\{ 9 - 2\} - 5}}\)
\(= \frac{{3 \times 7 + 5}}{{3 \times 7 - 5}}\)
\(= \frac{{21 + 5}}{{21 - 5}}\)
\(= \frac{{26}}{{16}}\)
\(= \frac{{13}}{8}\)
\(= 1\frac{5}{8}\) (উত্তর)
\(\therefore\) নির্ণেয় মান \(= 1\frac{5}{8}\)
(iv) \(\frac{{{a^3} + {b^3}}}{{{a^3} - {b^3}}}\)
সমাধান \(\frac{{{a^3} + {b^3}}}{{{a^3} - {b^3}}}\)
\(= \frac{{{{(a + b)}^3} - 3ab(a + b)}}{{{{(a - b)}^3} + 3ab(a - b)}}\)
\( = \frac{{{{(3)}^3} - 3 \times 1 \times 3}}{{{{(\sqrt 5 )}^3} + 3 \times 1 \times \sqrt 5 }}\)\([\because a+b=3, a b=1]\)
\(= \frac{{27 - 9}}{{5\sqrt 5 + 3\sqrt 5 }}\)
\(= \frac{{18}}{{8\sqrt 5 }}\)
\(= \frac{9}{{4\sqrt 5 }}\)
\(= \frac{{9 \times \sqrt 5 }}{{4\sqrt 5 \times \sqrt 5 }}\)
\(= \frac{{9\sqrt 5 }}{{20}}\) \(\therefore\) নির্ণেয় মান \(= \frac{{9\sqrt 5 }}{{20}}\)
7. যদি \(x = 2 + \sqrt 3 ,y = 2 - \sqrt 3 \) হয়, তবে নিম্নলিখিতগুলির সরলতম মান নির্ণয় করি
(a) \({\rm { (i) }} x - \frac{1}{x}\) \({\rm{ (ii) }}{y^2} + \frac{1}{{{y^2}}}\) \({\rm{ (iii) }}{x^3} - \frac{1}{{{x^3}}}\) \({\rm{(iv) }}xy + \frac{1}{{xy}}\)
প্রথমে x এবং y-এর মান নির্ণয় করে পাওয়া যায়।
\(x = 2 + \sqrt 3 \)
\(\frac{1}{x} = \frac{1}{{2 + \sqrt 3 }}\)
\(= \frac{{2 - \sqrt 3 }}{{(2 + \sqrt 3 )(2 - \sqrt 3 )}}\)
\(=\frac{(2-\sqrt{3})}{(2)^{2}-(\sqrt{3})^{2}}\)
\(= \frac{{2 - \sqrt 3 }}{{4 - 3}}\)
\(=\frac{2-\sqrt{3}}{1}\)
\( = 2 - \sqrt 3 \)
আবার, যেহেতু \(y = 2 - \sqrt 3 \)
\(\frac{1}{y}\) \(= \frac{1}{{2 - \sqrt 3 }}\)
\(= \frac{{2 + \sqrt 3 }}{{(2 - \sqrt 3 )(2 + \sqrt 3 )}}\)
\(=\frac{2+\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}\)
\(= \frac{{2 + \sqrt 3 }}{{4 - 3}}\)
\(=\frac{2+\sqrt{3}}{1}\)
\(= 2 + \sqrt 3 \)
a. (i) \(x - \frac{1}{x}\)
\(= (2 + \sqrt 3 ) - (2 - \sqrt 3 )\)
\(= 2 + \sqrt 3 - 2 + \sqrt 3 \)
\(= 2\sqrt 3 \)
\(\therefore\) নির্ণেয় মান \(= 2\sqrt 3 \)
(ii) \({y^2} + \frac{1}{{{y^2}}}\)
\(= {y^2} + \frac{1}{{{y^2}}}\)
\(= {y^2} + 2 \cdot y \cdot \frac{1}{y} + \frac{1}{{{y^2}}} - 2 \cdot y \cdot \frac{1}{y}\)
\( = {\left( {y + \frac{1}{y}} \right)^2} – 2\)
\(= {\{ (2 - \sqrt 3 ) + (2 + \sqrt 3 )\} ^2} - 2\)\(\left[\because y=2-\sqrt{3}, \frac{1}{y}=2+\sqrt{3}\right]\)
\( = {(4)^2} – 2\)
\( = 16 - 2 = 14\)
\(\therefore\) নির্ণেয় মান \(=14\)
(iii) \({x^3} - \frac{1}{{{x^3}}}\)
\(= {(x)^3} - {\left( {\frac{1}{x}} \right)^3}\)
\(= {\left( {x - \frac{1}{x}} \right)^3} + 3x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right)\)
\( = {\{ 2 + \sqrt 3 - (2 - \sqrt 3 )\} ^3} + 3 \times 1 \times\{ (2 + \sqrt 3 ) - (2 - \sqrt 3 )\} \) \(\left[\begin{array}{r}\because x=2+\sqrt{3} \\ \frac{1}{x}=2-\sqrt{3}\end{array}\right]\)
\( = {(2\sqrt 3 )^3} + 3 \cdot 2\sqrt 3 \)
\( = 8 \cdot 3\sqrt 3 + 6\sqrt 3\)
\(= 24\sqrt 3 + 6\sqrt 3 \)
\( = 30\sqrt 3 \)
\(\therefore\) নির্ণেয় মান \( = 30\sqrt 3 \)
(iv) \(xy = (2 + \sqrt 3 )(2 - \sqrt 3 )\)
\(= {(2)^2} - {(\sqrt 3 )^2}\)
\( = 4 - 3 = 1\)
\(\therefore xy + \frac{1}{{xy}} = 1 + \frac{1}{1} = 2\) (উত্তর)
\(\therefore\) নির্ণেয় মান \( =2 \)
\(x = 2 + \sqrt 3 \)
\(\frac{1}{x} = \frac{1}{{2 + \sqrt 3 }}\)
\(= \frac{{2 - \sqrt 3 }}{{(2 + \sqrt 3 )(2 - \sqrt 3 )}}\)
\(=\frac{(2-\sqrt{3})}{(2)^{2}-(\sqrt{3})^{2}}\)
\(= \frac{{2 - \sqrt 3 }}{{4 - 3}}\)
\(=\frac{2-\sqrt{3}}{1}\)
\( = 2 - \sqrt 3 \)
আবার, যেহেতু \(y = 2 - \sqrt 3 \)
\(\frac{1}{y}\) \(= \frac{1}{{2 - \sqrt 3 }}\)
\(= \frac{{2 + \sqrt 3 }}{{(2 - \sqrt 3 )(2 + \sqrt 3 )}}\)
\(=\frac{2+\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}\)
\(= \frac{{2 + \sqrt 3 }}{{4 - 3}}\)
\(=\frac{2+\sqrt{3}}{1}\)
\(= 2 + \sqrt 3 \)
a. (i) \(x - \frac{1}{x}\)
\(= (2 + \sqrt 3 ) - (2 - \sqrt 3 )\)
\(= 2 + \sqrt 3 - 2 + \sqrt 3 \)
\(= 2\sqrt 3 \)
\(\therefore\) নির্ণেয় মান \(= 2\sqrt 3 \)
(ii) \({y^2} + \frac{1}{{{y^2}}}\)
\(= {y^2} + \frac{1}{{{y^2}}}\)
\(= {y^2} + 2 \cdot y \cdot \frac{1}{y} + \frac{1}{{{y^2}}} - 2 \cdot y \cdot \frac{1}{y}\)
\( = {\left( {y + \frac{1}{y}} \right)^2} – 2\)
\(= {\{ (2 - \sqrt 3 ) + (2 + \sqrt 3 )\} ^2} - 2\)\(\left[\because y=2-\sqrt{3}, \frac{1}{y}=2+\sqrt{3}\right]\)
\( = {(4)^2} – 2\)
\( = 16 - 2 = 14\)
\(\therefore\) নির্ণেয় মান \(=14\)
(iii) \({x^3} - \frac{1}{{{x^3}}}\)
\(= {(x)^3} - {\left( {\frac{1}{x}} \right)^3}\)
\(= {\left( {x - \frac{1}{x}} \right)^3} + 3x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right)\)
\( = {\{ 2 + \sqrt 3 - (2 - \sqrt 3 )\} ^3} + 3 \times 1 \times\{ (2 + \sqrt 3 ) - (2 - \sqrt 3 )\} \) \(\left[\begin{array}{r}\because x=2+\sqrt{3} \\ \frac{1}{x}=2-\sqrt{3}\end{array}\right]\)
\( = {(2\sqrt 3 )^3} + 3 \cdot 2\sqrt 3 \)
\( = 8 \cdot 3\sqrt 3 + 6\sqrt 3\)
\(= 24\sqrt 3 + 6\sqrt 3 \)
\( = 30\sqrt 3 \)
\(\therefore\) নির্ণেয় মান \( = 30\sqrt 3 \)
(iv) \(xy = (2 + \sqrt 3 )(2 - \sqrt 3 )\)
\(= {(2)^2} - {(\sqrt 3 )^2}\)
\( = 4 - 3 = 1\)
\(\therefore xy + \frac{1}{{xy}} = 1 + \frac{1}{1} = 2\) (উত্তর)
\(\therefore\) নির্ণেয় মান \( =2 \)
(b) \(3{x^2} - 5xy + 3{y^2}\)
\(3{x^2} - 5xy + 3{y^2}\)
\( = 3{x^2} - 6xy + 3{y^2} + xy\)
\( = 3\left( {{x^2} - 2xy + {y^2}} \right) + xy\)
\( = 3{(x - y)^2} + xy\)
\( = 3{\{ (2 + \sqrt 3 ) - (2 - \sqrt 3 )\} ^2} + (2 + \sqrt 3 )(2 - \sqrt 3 )\) \(\left[\begin{array}{r}\because x=2+\sqrt{3} \\ y=2-\sqrt{3}\end{array}\right]\)
\( = 3{(2\sqrt 3 )^2} + \left\{ {{{(2)}^2} - {{(\sqrt 3 )}^2}} \right\}\)
\( = 3 \times 4 \times 3 + (4 - 3)\)
\( = 36 + 1 = 37\)
\(\therefore\) নির্ণেয় মান \(=37 \)
\( = 3{x^2} - 6xy + 3{y^2} + xy\)
\( = 3\left( {{x^2} - 2xy + {y^2}} \right) + xy\)
\( = 3{(x - y)^2} + xy\)
\( = 3{\{ (2 + \sqrt 3 ) - (2 - \sqrt 3 )\} ^2} + (2 + \sqrt 3 )(2 - \sqrt 3 )\) \(\left[\begin{array}{r}\because x=2+\sqrt{3} \\ y=2-\sqrt{3}\end{array}\right]\)
\( = 3{(2\sqrt 3 )^2} + \left\{ {{{(2)}^2} - {{(\sqrt 3 )}^2}} \right\}\)
\( = 3 \times 4 \times 3 + (4 - 3)\)
\( = 36 + 1 = 37\)
\(\therefore\) নির্ণেয় মান \(=37 \)
Koshe Dekhi 9.3 Class 10|দ্বিঘাত করণী কষে দেখি 9.3|গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3|WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Surds Solution in Bengali
8. \(x = \frac{{\sqrt 7 + \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }}\) এবং \(xy = 1\) হলে, দেখাই যে, \(\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = \frac{{12}}{{11}}\)
যেহেতু \(x = \frac{{\sqrt 7 + \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }}\) এবং xy = 1
\(\therefore y = \frac{1}{x} = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }}\)
এখন \(x + y = \frac{{\sqrt 7 + \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }} + \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }}\)
\( = \frac{{{{(\sqrt 7 + \sqrt 3 )}^2} + {{(\sqrt 7 - \sqrt 3 )}^2}}}{{(\sqrt 7 - \sqrt 3 )(\sqrt 7 + \sqrt 3 )}}\)
\(=\frac{(\sqrt{7})^{2}+2 \times \sqrt{7} \times \sqrt{3}+(\sqrt{3})^{2}+(\sqrt{7})^{2}-2 \times \sqrt{7} \times \sqrt{3}+(\sqrt{3})^{2}}{(\sqrt{7})^{2}-(\sqrt{3})^{2}}\)
\( = \frac{{(7 + 3 + 2\sqrt 7 \times \sqrt 3 ) + (7 + 3 - 2 \times \sqrt 7 \times \sqrt 3 )}}{{{{(\sqrt 7 )}^2} - {{(\sqrt 3 )}^2}}}\)
\(= \frac{{20}}{{7 - 3}}\)
\(= \frac{{20}}{4}\)
\( = 5\)
এখন \(\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}}\)
\(=\frac{{{x^2} + 2xy + {y^2}}-xy}{{{x^2} +2 xy + {y^2}}-3xy}\)
\(= \frac{{{{(x + y)}^2} - xy}}{{{{(x + y)}^2} - 3xy}}\)
\(= \frac{{{{( 5 )}^2} - 1}}{{{{(5)}^2} - 3.1}}\) \([\because x+y=5, x y=1]\)
\( = \frac{{25 - 1}}{{25 - 3}}\)
\(= \frac{{24}}{{22}}\)
\(= \frac{{12}}{{11}}\) (প্রমাণিত)
\(\therefore y = \frac{1}{x} = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }}\)
এখন \(x + y = \frac{{\sqrt 7 + \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }} + \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }}\)
\( = \frac{{{{(\sqrt 7 + \sqrt 3 )}^2} + {{(\sqrt 7 - \sqrt 3 )}^2}}}{{(\sqrt 7 - \sqrt 3 )(\sqrt 7 + \sqrt 3 )}}\)
\(=\frac{(\sqrt{7})^{2}+2 \times \sqrt{7} \times \sqrt{3}+(\sqrt{3})^{2}+(\sqrt{7})^{2}-2 \times \sqrt{7} \times \sqrt{3}+(\sqrt{3})^{2}}{(\sqrt{7})^{2}-(\sqrt{3})^{2}}\)
\( = \frac{{(7 + 3 + 2\sqrt 7 \times \sqrt 3 ) + (7 + 3 - 2 \times \sqrt 7 \times \sqrt 3 )}}{{{{(\sqrt 7 )}^2} - {{(\sqrt 3 )}^2}}}\)
\(= \frac{{20}}{{7 - 3}}\)
\(= \frac{{20}}{4}\)
\( = 5\)
এখন \(\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}}\)
\(=\frac{{{x^2} + 2xy + {y^2}}-xy}{{{x^2} +2 xy + {y^2}}-3xy}\)
\(= \frac{{{{(x + y)}^2} - xy}}{{{{(x + y)}^2} - 3xy}}\)
\(= \frac{{{{( 5 )}^2} - 1}}{{{{(5)}^2} - 3.1}}\) \([\because x+y=5, x y=1]\)
\( = \frac{{25 - 1}}{{25 - 3}}\)
\(= \frac{{24}}{{22}}\)
\(= \frac{{12}}{{11}}\) (প্রমাণিত)
9.
\((\sqrt 7 + 1)\) এবং \((\sqrt 5 + \sqrt 3 )\) এর মধ্যে কোনটি বড়ো লিখি।
\((\sqrt{7}+1)^{2}=7+1+2 \sqrt{7}=8+2 \sqrt{7}=8+2 \sqrt{7}\)
\((\sqrt{5}+\sqrt{3})^{2}=5+3+2 \sqrt{5 \times 3}=8+2 \sqrt{15}\)
\(\therefore(\sqrt{5}+\sqrt{3})^{2}>(\sqrt{7}+1)^{2}[\because \sqrt{15}>\sqrt{7}]\)
\(\therefore \sqrt{5}+\sqrt{3}>\sqrt{7}+1\)
উত্তর : \(\sqrt{5}+\sqrt{3}\) বড়াে।
\((\sqrt{5}+\sqrt{3})^{2}=5+3+2 \sqrt{5 \times 3}=8+2 \sqrt{15}\)
\(\therefore(\sqrt{5}+\sqrt{3})^{2}>(\sqrt{7}+1)^{2}[\because \sqrt{15}>\sqrt{7}]\)
\(\therefore \sqrt{5}+\sqrt{3}>\sqrt{7}+1\)
উত্তর : \(\sqrt{5}+\sqrt{3}\) বড়াে।
অতিসংক্ষিপ্ত উত্তরধৰ্মী প্ৰশ্ন (V.S.A.)
(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)
(i) \(x = 2 + \sqrt 3 \) হলে \(x + \frac{1}{x}\) এর মান
\({\rm{(a) }}2\) \({\rm{ (b) }}2\sqrt 3 \) \({\rm{(c) }}4\) \({\rm{ (d) }}2 - \sqrt 3 \)
\(x=2+\sqrt{3}\)
\(\therefore \frac{1}{x}=\frac{1}{2+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}\)
\(=\frac{2-\sqrt{3}}{4-3}\)
\(=\frac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
\(\therefore \quad x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4\)
\(\therefore\) (c) উত্তরটি সঠিক।
\(\therefore \frac{1}{x}=\frac{1}{2+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}\)
\(=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}\)
\(=\frac{2-\sqrt{3}}{4-3}\)
\(=\frac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
\(\therefore \quad x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4\)
\(\therefore\) (c) উত্তরটি সঠিক।
(ii) যদি \(p + q = \sqrt {13} \) এবং \(p - q = \sqrt 5 \) হয়, তাহলে pq এর মান (a) 2 (b) 18 (c) 9 (d) 8
আমরা জানি, \(p q=\left(\frac{p+q}{2}\right)^{2}-\left(\frac{p-q}{2}\right)^{2}\)
\(=\left(\frac{\sqrt{13}}{2}\right)^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}\)\([\because p+q=\sqrt{13}, p-q=\sqrt{5}]\)
\(=\frac{13}{4}-\frac{5}{4}=\frac{13-5}{4}\)
\(=\frac{8}{4}=2\)
\(\therefore\) (a) উত্তরটি সঠিক।
\(=\left(\frac{\sqrt{13}}{2}\right)^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}\)\([\because p+q=\sqrt{13}, p-q=\sqrt{5}]\)
\(=\frac{13}{4}-\frac{5}{4}=\frac{13-5}{4}\)
\(=\frac{8}{4}=2\)
\(\therefore\) (a) উত্তরটি সঠিক।
(iii) যদি \(a + b = \sqrt 5 \) এর \(a - b = \sqrt 3 \) হয় তাহলে \(\left( {{a^2} + {b^2}} \right)\) এর মান
(a) 8 (b) 4 (c) 2 (d) 1
(a) 8 (b) 4 (c) 2 (d) 1
আমরা জানি,
\(a^{2}+b^{2}\)
\(=\frac{1}{2} \times 2\left(a^{2}+b^{2}\right)\)
\(=\frac{1}{2}\left\{(a+b)^{2}+(a-b)^{2}\right\}\)
\(=\frac{1}{2}\left\{(\sqrt{5})^{2}+(\sqrt{3})^{2}\right\}\)\([\because a+b=\sqrt{5}, a-b=\sqrt{3}]\)
\(=\frac{1}{2}(5+3)\)
\(=\frac{1}{2} \times 8=4\)
\(\therefore\) (b) উত্তরটি সঠিক।
\(a^{2}+b^{2}\)
\(=\frac{1}{2} \times 2\left(a^{2}+b^{2}\right)\)
\(=\frac{1}{2}\left\{(a+b)^{2}+(a-b)^{2}\right\}\)
\(=\frac{1}{2}\left\{(\sqrt{5})^{2}+(\sqrt{3})^{2}\right\}\)\([\because a+b=\sqrt{5}, a-b=\sqrt{3}]\)
\(=\frac{1}{2}(5+3)\)
\(=\frac{1}{2} \times 8=4\)
\(\therefore\) (b) উত্তরটি সঠিক।
(iv) \(\sqrt {125} \) থেকে \(\sqrt 5 \) বিয়োগ করলে বিয়োগফল হবে \({\rm{(a) }}\sqrt {80} {\rm{ (b) }}\sqrt {120} {\rm{ (c) }}\sqrt {100} \) (d) কোনটিই নয়।
\(\sqrt{125}-\sqrt{5}\)
\(=\sqrt{25 \times 5}-\sqrt{5}=\sqrt{25} \times \sqrt{5}-\sqrt{5}\)
\(=5 \sqrt{5}-\sqrt{5}\)
\(=\sqrt{5}(5-1)\)
\(=4 \sqrt{5}\)
\(=\sqrt{16 \times 5}\)
\( =\sqrt{80}\)
\(\therefore\) (a) উত্তরটি সঠিক।
\(=\sqrt{25 \times 5}-\sqrt{5}=\sqrt{25} \times \sqrt{5}-\sqrt{5}\)
\(=5 \sqrt{5}-\sqrt{5}\)
\(=\sqrt{5}(5-1)\)
\(=4 \sqrt{5}\)
\(=\sqrt{16 \times 5}\)
\( =\sqrt{80}\)
\(\therefore\) (a) উত্তরটি সঠিক।
(v) \({\rm{ }}(5 - \sqrt 3 )(\sqrt 3 - 1)(5 + \sqrt 3 )(\sqrt 3 + 1)\) -এর গুণফল
(a) 22 (b) 44 (c) 2 (d) 11
(a) 22 (b) 44 (c) 2 (d) 11
\((5-\sqrt{3})(\sqrt{3}-1)(5+\sqrt{3})(\sqrt{3}+1)\)
\(=\{(5-\sqrt{3})(5+\sqrt{3})\}\{(\sqrt{3}-1)(\sqrt{3}+1)\}\)
\(=\left\{(\sqrt{5})^{2}-(\sqrt{3})^{2}\right\}\left\{(\sqrt{3})^{2}-(1)^{2}\right\}\)
\(=(25-3)(3-1)\)
\(=22 \times 2=44\)
(b) উত্তরটি সঠিক।
\(=\{(5-\sqrt{3})(5+\sqrt{3})\}\{(\sqrt{3}-1)(\sqrt{3}+1)\}\)
\(=\left\{(\sqrt{5})^{2}-(\sqrt{3})^{2}\right\}\left\{(\sqrt{3})^{2}-(1)^{2}\right\}\)
\(=(25-3)(3-1)\)
\(=22 \times 2=44\)
(b) উত্তরটি সঠিক।
(B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি
(i) \(\sqrt {75} \) এবং \(\sqrt {147} \) সদৃশ করণী
\(\sqrt{75}=\sqrt{25 \times 3} \)
\( =\sqrt{25} \times \sqrt{3}=5 \sqrt{3}\)
\(\sqrt{147}=\sqrt{49 \times 3}=\sqrt{49} \times \sqrt{3}=7 \sqrt{3}\)
\(\therefore \sqrt{75}\) এবং \(\sqrt{147}\) দুটি সদৃশ করণী।
\(\therefore\) বিবৃতিটি সত্য।
\( =\sqrt{25} \times \sqrt{3}=5 \sqrt{3}\)
\(\sqrt{147}=\sqrt{49 \times 3}=\sqrt{49} \times \sqrt{3}=7 \sqrt{3}\)
\(\therefore \sqrt{75}\) এবং \(\sqrt{147}\) দুটি সদৃশ করণী।
\(\therefore\) বিবৃতিটি সত্য।
(ii) \(\sqrt \pi \) একটি দ্বিঘাত করণী।
মিথ্যা, কারণ \(\sqrt{\pi}\) একটি তুরীয় (Transcendental) অমূলদ সংখ্যা।
(C) শূন্যস্থান পূরণ করি
(i) \(5\sqrt {11} \) একটি______সংখ্যা। (মূলদ/ অমূলদ)
অমূলদ
(ii) \((\sqrt 3 - 5)\) এর অনুবন্ধী করণী ______।
\((\sqrt{3}-5)(-\sqrt{3}-5)\)
\(=-(\sqrt{3}-5)(\sqrt{3}+5)\)
\(=-\left\{(\sqrt{3})^{2}-(5)^{2}\right\}\)
\(=-(3-25)\)
\(=-(-22)=22\) যা একটি মূলদ সংখ্যা।
\(\therefore\) নির্ণেয় অনুবন্ধী করণী \((-\sqrt{3}-5)\)।
[আবার, \((\sqrt{3}-5)(\sqrt{3}+5)\)
\(=(\sqrt{3})^{2}-(5)^{2}=3-25=-22\), যা একটি মূলদ সংখ্যা।
\(\therefore(\sqrt{3}-5)\)-এর অনুবন্ধী করণী \((\sqrt{3}+5)\)]
\(=-(\sqrt{3}-5)(\sqrt{3}+5)\)
\(=-\left\{(\sqrt{3})^{2}-(5)^{2}\right\}\)
\(=-(3-25)\)
\(=-(-22)=22\) যা একটি মূলদ সংখ্যা।
\(\therefore\) নির্ণেয় অনুবন্ধী করণী \((-\sqrt{3}-5)\)।
[আবার, \((\sqrt{3}-5)(\sqrt{3}+5)\)
\(=(\sqrt{3})^{2}-(5)^{2}=3-25=-22\), যা একটি মূলদ সংখ্যা।
\(\therefore(\sqrt{3}-5)\)-এর অনুবন্ধী করণী \((\sqrt{3}+5)\)]
(iii) দুটি দ্বিঘাত করণীর যোগফল ও গুণফল একটি মূলদ সংখ্যা হলে করণীদ্বয় ______ করণী।
অনুবন্ধী।
সংক্ষিপ্তধমী উত্তর প্রশ্ন (S.A.)
11.
(i) \(x = 3 + 2\sqrt 2 \) হলে \(x + \frac{1}{x}\) -এর মান লিখি।
\(x=3+2 \sqrt{2},\)
\(\therefore \frac{1}{x}=\frac{1}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}\)
\(=\frac{(3-2 \sqrt{2})}{(3+2 \sqrt{2})(3-2 \sqrt{2})}\)
\(=\frac{3-2 \sqrt{2}}{(3)^{2}-(2 \sqrt{2})^{2}}\)
\(=\frac{3-2 \sqrt{2}}{9-8}\)
\(=\frac{3-2 \sqrt{2}}{1}\)
\(=3-2 \sqrt{2}\)
\(\therefore x+\frac{1}{x}=3+2 \sqrt{2}+3-2 \sqrt{2}=6\)
\(\therefore\) নির্ণেয় গুণফল \(= 6\)।
\(\therefore \frac{1}{x}=\frac{1}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}\)
\(=\frac{(3-2 \sqrt{2})}{(3+2 \sqrt{2})(3-2 \sqrt{2})}\)
\(=\frac{3-2 \sqrt{2}}{(3)^{2}-(2 \sqrt{2})^{2}}\)
\(=\frac{3-2 \sqrt{2}}{9-8}\)
\(=\frac{3-2 \sqrt{2}}{1}\)
\(=3-2 \sqrt{2}\)
\(\therefore x+\frac{1}{x}=3+2 \sqrt{2}+3-2 \sqrt{2}=6\)
\(\therefore\) নির্ণেয় গুণফল \(= 6\)।
\({\rm{(ii) }}(\sqrt {15} + \sqrt 3 )\) এবং \((\sqrt {10} + \sqrt 8 )\) এর মধ্যে কোনটি বড়ো লিখি।
\((\sqrt{15}+\sqrt{3})^{2}\)
\(=(\sqrt{15})^{2}+2 \times \sqrt{15} \times \sqrt{3}+(\sqrt{3})^{2}\)
\(=15+2 \times \sqrt{15 \times 3}+3\)
\(=18+2 \times \sqrt{45}\)
আবার, \((\sqrt{10}+\sqrt{8})^{2}\)
\(=(\sqrt{10})^{2}+2 \times \sqrt{10} \times \sqrt{8}+(\sqrt{8})^{2}\)
\(=10+2 \times \sqrt{10 \times 8}+8=18+2 \sqrt{80}\)
\(\because \sqrt{80}>\sqrt{45}, \)
\(\therefore(\sqrt{15}+\sqrt{3})^{2}<(\sqrt{10}+\sqrt{8})^{2}\)
\(\Rightarrow(\sqrt{10}+\sqrt{8})>(\sqrt{15}+\sqrt{3})\)
\(\therefore (\sqrt{10}+\sqrt{8})\)করণীটি বড়াে।
\(=(\sqrt{15})^{2}+2 \times \sqrt{15} \times \sqrt{3}+(\sqrt{3})^{2}\)
\(=15+2 \times \sqrt{15 \times 3}+3\)
\(=18+2 \times \sqrt{45}\)
আবার, \((\sqrt{10}+\sqrt{8})^{2}\)
\(=(\sqrt{10})^{2}+2 \times \sqrt{10} \times \sqrt{8}+(\sqrt{8})^{2}\)
\(=10+2 \times \sqrt{10 \times 8}+8=18+2 \sqrt{80}\)
\(\because \sqrt{80}>\sqrt{45}, \)
\(\therefore(\sqrt{15}+\sqrt{3})^{2}<(\sqrt{10}+\sqrt{8})^{2}\)
\(\Rightarrow(\sqrt{10}+\sqrt{8})>(\sqrt{15}+\sqrt{3})\)
\(\therefore (\sqrt{10}+\sqrt{8})\)করণীটি বড়াে।
Koshe Dekhi 9.3 Class 10|দ্বিঘাত করণী কষে দেখি 9.3|গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3|WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Surds Solution in Bengali
(iii) দুটি মিশ্র দ্বিঘাত করণী লিখি যাদের গুণফল একটি মূলদ সংখ্যা।
ধরি, একটি সংখ্যা \(=7 \sqrt{3}-5 \sqrt{2}\)
\(\therefore\) অপরটি \(=7 \sqrt{3}+5 \sqrt{2}\) কারণ এটি অনুবন্ধী করণী এবং আমরা জানি, দুটি পরস্পর অনুবন্ধী করণীর গুণফল একটি মূলদ সংখ্যা। অথবা, \(a+\sqrt{b}\) এবং \(a-\sqrt{b}\) যেখানে \(a\) মূল্পদ সংখ্যা এবং \(\sqrt{b}\) একটি শুদ্ধ করণী।
\(\therefore\) অপরটি \(=7 \sqrt{3}+5 \sqrt{2}\) কারণ এটি অনুবন্ধী করণী এবং আমরা জানি, দুটি পরস্পর অনুবন্ধী করণীর গুণফল একটি মূলদ সংখ্যা। অথবা, \(a+\sqrt{b}\) এবং \(a-\sqrt{b}\) যেখানে \(a\) মূল্পদ সংখ্যা এবং \(\sqrt{b}\) একটি শুদ্ধ করণী।
(iv) \(\sqrt {72} \) থেকে কত বিয়োগ করলে \(\sqrt {32} \) হবে তা লিখি।
\(\sqrt{72}-\sqrt{32}\)
\(=\sqrt{36 \times 2}-\sqrt{16 \times 2}\)
\( =\sqrt{36} \times \sqrt{2}-\sqrt{16} \times \sqrt{2}\)
\(=6 \sqrt{2}-4 \sqrt{2}\)
\(=(6-4) \sqrt{2}\) \(=2 \sqrt{2}\)
\(\therefore 2 \sqrt{2}\) বিয়ােগ করতে হবে।
\(=\sqrt{36 \times 2}-\sqrt{16 \times 2}\)
\( =\sqrt{36} \times \sqrt{2}-\sqrt{16} \times \sqrt{2}\)
\(=6 \sqrt{2}-4 \sqrt{2}\)
\(=(6-4) \sqrt{2}\) \(=2 \sqrt{2}\)
\(\therefore 2 \sqrt{2}\) বিয়ােগ করতে হবে।
(v) \(\left( {\frac{1}{{\sqrt 2 + 1}} + \frac{1}{{\sqrt 3 + \sqrt 2 }} + \frac{1}{{\sqrt 4 + \sqrt 3 }}} \right)\) এর সরলতম মান লিখি।
\(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\)
\(=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}}\)
\(=\frac{\sqrt{2}-1}{(\sqrt{2})^{2}-(1)^{2}}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4})^{2}-(\sqrt{3})^{2}}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}=-1+\sqrt{4}=-1+2=1\)
\(\therefore\) নির্ণেয় সরলতম মান \(= 1\)।
\(=\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}}\)
\(=\frac{\sqrt{2}-1}{(\sqrt{2})^{2}-(1)^{2}}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^{2}}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4})^{2}-(\sqrt{3})^{2}}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}=-1+\sqrt{4}=-1+2=1\)
\(\therefore\) নির্ণেয় সরলতম মান \(= 1\)।
Koshe Dekhi 9.3 Class 10|দ্বিঘাত করণী কষে দেখি 9.3|গণিত প্রকাশ দশম শ্রেণী দ্বিঘাত করণী কষে দেখি -9.3|WBBSE Ganit Prakash Class 10(Ten) (X) Koshe Dekhi 9.3 (Exercise -9.3) Chapter 9 Surds Solution in Bengali
এই Page টি বা এই Website টির কোন প্রকার বিষয়বস্তু কপি করা বা সম্পাদনা করা নিষিদ্ধ। ভারতীয় Copywright আইন 1957 এর ধারা 63 অনুযায়ী, এই ফাইলটির সমস্ত অধিকার 'ছাত্র মিত্র Mathematics' অ্যাপ দ্বারা সংরক্ষিত। ছাত্র মিত্রের অনুমতি ছাড়া, এই Page টি বা এই Website টির কোন প্রকার বিষয়বস্তু কপি করা বা সম্পাদনা করা আইনত দন্ডনীয় অপরাধ। কেউ ছাত্র মিত্রের অনুমতি ছাড়া, এই Page টি বা এই Website টির কোন প্রকার বিষয়বস্তু কপি বা সম্পাদনা করলে ছাত্র মিত্র কতৃপক্ষ তার বিরুদ্ধে সকল প্রকার কঠোর আইনি পদক্ষেপ করবে।
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Class 8 : গণিত প্রভা (অষ্টম শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali.
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Class 6 : গণিত প্রভা (ষষ্ঠ শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali
Class 9 : গণিত প্রকাশ (নবম শ্রেণি) বইয়ের সমাধান Maths Solution WBBSE Bengali
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Class 8 : গণিত প্রভা (অষ্টম শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali.
Class 7 : গণিত প্রভা (সপ্তম শ্রেণি) বইয়ের সমস্ত সমাধান
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Class 10 : মাধ্যমিক গণিত প্রকাশ (দশম শ্রেণি) বইয়ের সমস্ত সমাধান Class 10 Maths Solution WBBSE Bengali
Class 6 : গণিত প্রভা (ষষ্ঠ শ্রেণি) বইয়ের সমস্ত সমাধান Maths solutions for WBBSE in Bengali
Class 9 : গণিত প্রকাশ (নবম শ্রেণি) বইয়ের সমাধান Maths Solution WBBSE Bengali
আজই Install করুন Chatra Mitra
