Question

Pls solve this question​

Answer 1

Answer

LHS = RHS

Given

$\bullet \rm \;\; \dfrac{(x^{-1}+\gamma^{-1})}{x^{-1}}+\dfrac{(x^{-1}+\gamma^{-1})}{\gamma^{-1}}$

To prove

$\bullet \;\; \rm \dfrac{(x+\gamma)^2}{x\gamma}$

Proof

$\rm LHS\\\\\rm \implies \dfrac{(x^{-1}+\gamma^{-1})}{x^{-1}}+\dfrac{(x^{-1}+\gamma^{-1})}{\gamma^{-1}}\\\\\implies \rm \dfrac{(\frac{1}{x}+\frac{1}{\gamma})}{\frac{1}{x}}+\dfrac{(\frac{1}{x}+\frac{1}{\gamma})}{\frac{1}{\gamma}}\\\\\implies \rm \dfrac{\frac{(\gamma +x)}{x \gamma}}{\frac{1}{x}}+ \dfrac{\frac{(\gamma +x)}{x \gamma}}{\frac{1}{\gamma}}\\\\\implies \rm \dfrac{x(\gamma +x)}{x\gamma}+\dfrac{\gamma(\gamma + x)}{x\gamma}\\\\\implies \rm \dfrac{x(\gamma +x)+\gamma (\gamma +x)}{x\gamma}$

$\implies \rm \dfrac{x\gamma +x^2+\gamma^2+\gamma x}{x\gamma}\\\\\implies \rm \dfrac{x^2+2x\gamma +\gamma^2}{x\gamma}\\\\\implies \rm \dfrac{(x+\gamma)^2}{x\gamma}\ \; [\; Since,(a+b)^2=a^2+2ab+b^2 \;]\\\\\implies \rm RHS\\\\\rm Hence\ proved$

Answer 2

QUESTION:-

➣ Prove:-

•ANSWER:-

➣ Given:-

$\dfrac{(x^{-1}+y^{-1})}{x^{-1}}+\dfrac{(x^{-1}+y^{-1})}{y^{-1}}$

➣ To Prove:-

$\dfrac{(x+y)^2}{xy}$

➣ Proof:-

• LHS :-

➣ $\dfrac{(x^{-1}+y^{-1})}{x^{-1}}+\dfrac{(x^{-1}+y^{-1})}{y^{-1}}$

➣ $\dfrac{(\frac{1}{x}+\frac{1}{y})}{\frac{1}{x}}+\dfrac{(\frac{1}{x}+\frac{1}{y})}{\frac{1}{y}}$

➣ $\dfrac{\frac{(y+x)}{x y}}{\frac{1}{x}}+ \dfrac{\frac{(y +x)}{x y}}{\frac{1}{y}}$

➣$\dfrac{x(y +x)}{xy}+\dfrac{y(y + x)}{xy}$

➣ $\dfrac{x(y +x)+y(y +x)}{xy}$

➣ $\dfrac{xy +x^2+y^2+y x}{xy}$

➣ $\dfrac{x^2+2xy +y^2}{xy}$

$\boxed{\bold{As \ we \ know \ the \ identity:-}} \\ </p><p></p><p>{\boxed{\bold(a+b)^2=a^2+2ab+b^2}}$

➣ $\dfrac{(x+y)^2}{xy}$

• RHS:-

➣ $\dfrac{(x+y)^2}{xy}$

• Therefore,

$\purple{LHS = RHS }$

Hence Proved....!