Simlify-1-x-1-x-1-x-1-x-2-1-x-1-x-1-x-1-x-2-

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

{(\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