Question

( x+1/x-1 - x-1/x+1 ) ÷ (x-1/x+1 + x+1/x-1) = 2x/x²+1 prove it​

Answer 1

Answer:

here answer for your question

Answer 2

Answer:

Simplified value of this equation is 2x / ( x^2 + 1 ).

Step-by-step explanation:

$\implies \bigg \{ \dfrac{x + 1}{x - 1} - \dfrac{x - 1}{x + 1} \bigg \} : \bigg \{\dfrac{x -1}{x + 1} + \ \dfrac{x + 1}{x - 1} \bigg \} \\ \\ \\ \implies \bigg \{ \dfrac{(x + 1) {}^{2} - (x - 1) {}^{2} }{(x - 1)(x + 1)} \bigg \} : \bigg \{ \dfrac{(x - 1) {}^{2} + (x + 1) {}^{2} }{(x + 1)(x - 1)} \bigg \}$

Using :

• ( a + b )^2 - ( a - b )^2 = a^2 + b^2 + 2ab - ( a^2 + b^2 - 2ab ) = 4ab
• ( a - b )^2 + ( a + b )^2 = a^2 + b^2 - 2ab + a^2 + b^2 + 2ab = 2( a^2 + b^2 )
• ( a + b )( a - b ) = a^2 - b^2

$\implies \bigg \{ \dfrac{4x}{x {}^{2} - 1} \bigg \} : \bigg \{ \dfrac{2(x {}^{2} + 1) {}^{} {}^{} }{x {}^{2} - 1 } \bigg \} \\ \\ \\ \implies \dfrac{4x}{2(x {}^{2} + 1) } \\ \\ \\ \implies \dfrac{2x}{x {}^{2} + 1}$

Hence,proved.