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

Simlify (((1+(√x))/( (√(1+x)))) − ((√(1+x))/(1+(√x))))^2 - (((1−(√x))/( (√(1+x)))) − ((√(1+x))/(1−(√x))))^2

$${Simlify} \\ $$$$\left(\frac{\mathrm{1}+\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}}}\:−\:\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{2}} -\:\left(\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}}}\:−\:\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}−\sqrt{{x}}}\right)^{\mathrm{2}} \\ $$

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 ))

$$\left(\underset{{a}} {\underbrace{\frac{\mathrm{1}+\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}}}\:}}−\:\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{2}} -\:\left(\underset{{b}} {\underbrace{\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}}}}}\:−\:\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}−\sqrt{{x}}}\right)^{\mathrm{2}} \\ $$$$\left({a}−\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} −\left({b}−\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} \\ $$$$=\left\{\left({a}−\frac{\mathrm{1}}{{a}}\right)−\left({b}−\frac{\mathrm{1}}{{b}}\right)\right\}\left\{\left({a}−\frac{\mathrm{1}}{{a}}\right)+\left({b}−\frac{\mathrm{1}}{{b}}\right)\right\} \\ $$$$=\left({a}−{b}−\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left({a}+{b}−\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right) \\ $$$$=\left({a}−{b}+\frac{{a}−{b}}{{ab}}\right)\left({a}+{b}−\frac{{a}+{b}}{{ab}}\right) \\ $$$$=\left({a}−{b}\right)\left({a}+{b}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{{ab}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{ab}}\right) \\ $$$$=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\:\mathrm{1}−\left(\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} \:\right) \\ $$$$\:\left\{\left(\frac{\mathrm{1}+\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{\mathrm{1}+{x}}}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left\{\mathrm{1}−\left(\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}−\sqrt{{x}}}\right)^{\mathrm{2}} \right\} \\ $$$$\:=\left\{\frac{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{{x}}\right)^{\mathrm{2}} }{\:\mathrm{1}+{x}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left\{\mathrm{1}−\left(\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{1}+{x}}}{\mathrm{1}−\sqrt{{x}}}\right)^{\mathrm{2}} \right\} \\ $$$$\:=\left\{\frac{\cancel{\mathrm{1}}+\cancel{{x}}+\mathrm{2}\sqrt{{x}}−\cancel{\mathrm{1}}−\cancel{{x}}+\mathrm{2}\sqrt{{x}}}{\:\mathrm{1}+{x}}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left\{\mathrm{1}−\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{4}\sqrt{{x}}}{\mathrm{1}+{x}}×\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}\sqrt{{x}}}{\mathrm{1}+{x}}×\frac{\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{2}{x}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}\sqrt{{x}}}{\mathrm{1}+{x}}×\frac{−\mathrm{4}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=−\frac{\mathrm{16}{x}\sqrt{{x}}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$

Commented by mathdanisur last updated on 21/Jul/21

Thankyou Sir, please note the answer too please if possible

$${Thankyou}\:{Sir},\:{please}\:{note}\:{the}\:{answer} \\ $$$${too}\:{please}\:{if}\:{possible} \\ $$

Commented by mathdanisur last updated on 22/Jul/21

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

Answered by liberty last updated on 22/Jul/21

(√x) = u ⇒(((1+u)/( (√(1+u^2 ))))−((√(1+u^2 ))/(1+u)))^2 −(((1−u)/( (√(1+u^2 ))))−((√(1+u^2 ))/(1−u)))^2 = (((1+u)/( (√(1+u^2 ))))−((√(1+u^2 ))/(1+u))+((1−u)/( (√(1+u^2 ))))−((√(1+u^2 ))/(1−u)) )(((1+u)/( (√(1+u^2 ))))−((√(1+u^2 ))/(1+u))−((1−u)/( (√(1+u^2 ))))+((√(1+u^2 ))/(1−u)) ) =((2/( (√(1+u^2 )))) −(√(1+u^2 )) ((1/(1+u))+(1/(1−u))))(((2u)/( (√(1+u^2 )))) +(√(1+u^2 ))((1/(1−u))−(1/(1+u)))) = ((2/( (√(1+u^2 ))))−((2(√(1+u^2 )))/(1−u^2 )) )(((2u)/( (√(1+u^2 ))))+((2u(√(1+u^2 )))/(1−u^2 )) ) = (((2−2u^2 −2−2u^2 )/((1−u^2 )(√(1+u^2 )))))(((2u−2u^3 +2u+2u^3 )/((1−u^2 )(√(1+u^2 )))) ) = (((−4u^2 )/((1−u^2 )(√(1+u^2 )))))(((4u)/((1−u^2 )(√(1+u^2 ))))) = ((−16u^3 )/((1−u^2 )^2 (1+u^2 ))) = ((−16x(√x))/((1−x)^2 (1+x))) .

$$\:\sqrt{{x}}\:=\:{u}\: \\ $$$$\Rightarrow\left(\frac{\mathrm{1}+{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}−\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}+{u}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}−{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}−\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}−{u}}\right)^{\mathrm{2}} = \\ $$$$\left(\frac{\mathrm{1}+{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}−\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}+{u}}+\frac{\mathrm{1}−{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}−\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}−{u}}\:\right)\left(\frac{\mathrm{1}+{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}−\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}+{u}}−\frac{\mathrm{1}−{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}+\frac{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}−{u}}\:\right) \\ $$$$=\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:−\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\:\left(\frac{\mathrm{1}}{\mathrm{1}+{u}}+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)\right)\left(\frac{\mathrm{2}{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}−\frac{\mathrm{1}}{\mathrm{1}+{u}}\right)\right) \\ $$$$=\:\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}−\frac{\mathrm{2}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}−{u}^{\mathrm{2}} }\:\right)\left(\frac{\mathrm{2}{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}+\frac{\mathrm{2}{u}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{\mathrm{1}−{u}^{\mathrm{2}} }\:\right) \\ $$$$=\:\left(\frac{\mathrm{2}−\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}−\mathrm{2}{u}^{\mathrm{2}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\right)\left(\frac{\mathrm{2}{u}−\mathrm{2}{u}^{\mathrm{3}} +\mathrm{2}{u}+\mathrm{2}{u}^{\mathrm{3}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:\right) \\ $$$$=\:\left(\frac{−\mathrm{4}{u}^{\mathrm{2}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\right)\left(\frac{\mathrm{4}{u}}{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\right) \\ $$$$=\:\frac{−\mathrm{16}{u}^{\mathrm{3}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\:\frac{−\mathrm{16}{x}\sqrt{{x}}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}\:. \\ $$

Commented by mathdanisur last updated on 22/Jul/21

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

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