Question

Plz solve these.... I'll mark u the brightest ​

Answer 1

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$\underline{\blacksquare\:\:\:\footnotesize{\red{SolutioNn:}}}$

$\footnotesize{x=\sqrt{3}-\sqrt{2}}$

$\footnotesize{\dfrac{1}{x}=\dfrac{1}{\sqrt{3}-\sqrt{2}}}$

$\footnotesize{\dfrac{1}{x}=\dfrac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}}$

$\footnotesize{\dfrac{1}{x}=\dfrac{(\sqrt{3}+\sqrt{2})}{\sqrt{3}^2-\sqrt{2}^2}}$

$\footnotesize{\dfrac{1}{x}=\dfrac{(\sqrt{3}+\sqrt{2})}{3-2}}$

$\footnotesize{\dfrac{1}{x}=\dfrac{(\sqrt{3}+\sqrt{2})}{1}}$

$\footnotesize{\dfrac{1}{x}=(\sqrt{3}+\sqrt{2})}$

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$\underline{\footnotesize{\red{Q-1):}}}$

$\footnotesize{\therefore\:x+\dfrac{1}{x}}$

$\footnotesize{=\sqrt{3}-\cancel {\sqrt{2}}+\sqrt{3}+\cancel {=\sqrt{2}}}$

$\footnotesize{=2\sqrt{3}}$

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$\underline{\footnotesize{\red{Q-2):}}}$

$\footnotesize{\therefore\:x^2+\dfrac{1}{x^2}}$

$\footnotesize{=(x+\dfrac{1}{x})^2-2\times \cancel{x}\times\dfrac{1}{\cancel{x}}}$

$\footnotesize{=(2\sqrt{3})^2-2}$

$\footnotesize{=12-2}$

$\footnotesize{=10}$

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$\underline{\footnotesize{\red{Q-3):}}}$

$\footnotesize{\therefore\:x^3+\dfrac{1}{x^3}}$

$\footnotesize{= (x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-\cancel{x}\dfrac{1}{\cancel{x^2}})}$

$\footnotesize{= (2\sqrt{3})(10-1)}$

$\footnotesize{= (2\sqrt{3})(9)}$

$\footnotesize{= 18\sqrt{3}}$

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$\underline{\footnotesize{\red{Q-4):}}}$

$\footnotesize{\therefore\:(x^3+\dfrac{1}{x^3})-3(x^2+\dfrac{1}{x^2})+(x+\dfrac{1}{x})}$

$\footnotesize{=(18\sqrt{3})-3(10)+(2\sqrt {3})}$

$\footnotesize{=16\sqrt{3}-30}$

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