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Question Number 147557 by mathdanisur last updated on 21/Jul/21

$S i m l i f y$ ${(\frac{1 + \sqrt{x}}{\sqrt{1 + x}} - \frac{\sqrt{1 + x}}{1 + \sqrt{x}})}^{2} - {(\frac{1 - \sqrt{x}}{\sqrt{1 + x}} - \frac{\sqrt{1 + x}}{1 - \sqrt{x}})}^{2}$
	Answered by Rasheed.Sindhi last updated on 21/Jul/21
	$(((1+(√x))/( (√(1+x)))) _(a) − ((√(1+x))/(1+(√x))))^2 - (((1−(√x))/( (√(1+x))))_(b) − ((√(1+x))/(1−(√x))))^2 (a−(1/a))^2 −(b−(1/b))^2 ={(a−(1/a))−(b−(1/b))}{(a−(1/a))+(b−(1/b))} =(a−b−(1/a)+(1/b))(a+b−(1/a)−(1/b)) =(a−b+((a−b)/(ab)))(a+b−((a+b)/(ab))) =(a−b)(a+b)(1+(1/(ab)))(1−(1/(ab))) =(a^2 −b^2 )( 1−((1/a))^2 ((1/b))^2 ) {(((1+(√x))/( (√(1+x)))))^2 −(((1−(√x))/( (√(1+x)))))^2 } ×{1−(((√(1+x))/(1+(√x))))^2 (((√(1+x))/(1−(√x))))^2 } ={(((1+(√x))^2 −(1−(√x))^2 )/( 1+x))} ×{1−(((√(1+x))/(1+(√x))))^2 (((√(1+x))/(1−(√x))))^2 } ={((1+x+2(√x)−1−x+2(√x))/( 1+x))} ×{1−(((1+x)^2 )/((1−x)^2 ))} =((4(√x))/(1+x))×(((1−x)^2 −(1+x)^2 )/((1−x)^2 )) =((4(√x))/(1+x))×((1−2x+x^2 −1−2x−x^2 )/((1−x)^2 )) =((4(√x))/(1+x))×((−4x)/((1−x)^2 ))=−((16x(√x))/((1+x)(1−x)^2 ))$
	${(\underset{a}{\underset{⏟}{\frac{1 + \sqrt{x}}{\sqrt{1 + x}}}} - \frac{\sqrt{1 + x}}{1 + \sqrt{x}})}^{2} - {(\underset{b}{\underset{⏟}{\frac{1 - \sqrt{x}}{\sqrt{1 + x}}}} - \frac{\sqrt{1 + x}}{1 - \sqrt{x}})}^{2}$ ${(a - \frac{1}{a})}^{2} - {(b - \frac{1}{b})}^{2}$ $= {(a - \frac{1}{a}) - (b - \frac{1}{b})} {(a - \frac{1}{a}) + (b - \frac{1}{b})}$ $= (a - b - \frac{1}{a} + \frac{1}{b}) (a + b - \frac{1}{a} - \frac{1}{b})$ $= (a - b + \frac{a - b}{a b}) (a + b - \frac{a + b}{a b})$ $= (a - b) (a + b) (1 + \frac{1}{a b}) (1 - \frac{1}{a b})$ $= (a^{2} - b^{2}) (1 - {(\frac{1}{a})}^{2} {(\frac{1}{b})}^{2})$ ${{(\frac{1 + \sqrt{x}}{\sqrt{1 + x}})}^{2} - {(\frac{1 - \sqrt{x}}{\sqrt{1 + x}})}^{2}}$ $\times {1 - {(\frac{\sqrt{1 + x}}{1 + \sqrt{x}})}^{2} {(\frac{\sqrt{1 + x}}{1 - \sqrt{x}})}^{2}}$ $= {\frac{{(1 + \sqrt{x})}^{2} - {(1 - \sqrt{x})}^{2}}{1 + x}}$ $\times {1 - {(\frac{\sqrt{1 + x}}{1 + \sqrt{x}})}^{2} {(\frac{\sqrt{1 + x}}{1 - \sqrt{x}})}^{2}}$ $= {\frac{1 + x + 2 \sqrt{x} - 1 - x + 2 \sqrt{x}}{1 + x}}$ $\times {1 - \frac{{(1 + x)}^{2}}{{(1 - x)}^{2}}}$ $= \frac{4 \sqrt{x}}{1 + x} \times \frac{{(1 - x)}^{2} - {(1 + x)}^{2}}{{(1 - x)}^{2}}$ $= \frac{4 \sqrt{x}}{1 + x} \times \frac{1 - 2 x + x^{2} - 1 - 2 x - x^{2}}{{(1 - x)}^{2}}$ $= \frac{4 \sqrt{x}}{1 + x} \times \frac{- 4 x}{{(1 - x)}^{2}} = - \frac{16 x \sqrt{x}}{(1 + x) {(1 - x)}^{2}}$
	Commented by mathdanisur last updated on 21/Jul/21

	$T h a n k y o u S i r, p l e a s e n o t e t h e a n s w e r$ $t o o p l e a s e i f p o s s i b l e$
	Commented by mathdanisur last updated on 22/Jul/21

	$t h a n k y o u S i r$
	Answered by liberty last updated on 22/Jul/21

	$\sqrt{x} = u$ $\Rightarrow {(\frac{1 + u}{\sqrt{1 + u^{2}}} - \frac{\sqrt{1 + u^{2}}}{1 + u})}^{2} - {(\frac{1 - u}{\sqrt{1 + u^{2}}} - \frac{\sqrt{1 + u^{2}}}{1 - u})}^{2} =$ $(\frac{1 + u}{\sqrt{1 + u^{2}}} - \frac{\sqrt{1 + u^{2}}}{1 + u} + \frac{1 - u}{\sqrt{1 + u^{2}}} - \frac{\sqrt{1 + u^{2}}}{1 - u}) (\frac{1 + u}{\sqrt{1 + u^{2}}} - \frac{\sqrt{1 + u^{2}}}{1 + u} - \frac{1 - u}{\sqrt{1 + u^{2}}} + \frac{\sqrt{1 + u^{2}}}{1 - u})$ $= (\frac{2}{\sqrt{1 + u^{2}}} - \sqrt{1 + u^{2}} (\frac{1}{1 + u} + \frac{1}{1 - u})) (\frac{2 u}{\sqrt{1 + u^{2}}} + \sqrt{1 + u^{2}} (\frac{1}{1 - u} - \frac{1}{1 + u}))$ $= (\frac{2}{\sqrt{1 + u^{2}}} - \frac{2 \sqrt{1 + u^{2}}}{1 - u^{2}}) (\frac{2 u}{\sqrt{1 + u^{2}}} + \frac{2 u \sqrt{1 + u^{2}}}{1 - u^{2}})$ $= (\frac{2 - 2 u^{2} - 2 - 2 u^{2}}{(1 - u^{2}) \sqrt{1 + u^{2}}}) (\frac{2 u - 2 u^{3} + 2 u + 2 u^{3}}{(1 - u^{2}) \sqrt{1 + u^{2}}})$ $= (\frac{- 4 u^{2}}{(1 - u^{2}) \sqrt{1 + u^{2}}}) (\frac{4 u}{(1 - u^{2}) \sqrt{1 + u^{2}}})$ $= \frac{- 16 u^{3}}{{(1 - u^{2})}^{2} (1 + u^{2})}$ $= \frac{- 16 x \sqrt{x}}{{(1 - x)}^{2} (1 + x)} .$
	Commented by mathdanisur last updated on 22/Jul/21

	$t h a n k y o u S i r$
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