Question

Como se resolve essa equação?
$\frac{\frac{2}{x-1} + \frac{x-1}{x+1} }{1+ \frac{x+1}{x-1} }$

Answer 1

$\dfrac{ \dfrac{2}{x - 1} + \dfrac{x - 1}{x + 1}}{1 + \dfrac{x + 1}{x - 1} } \\ \\ \\ \\ \dfrac{ \dfrac{x^2 + 3}{(x + 1)(x - 1)}}{\dfrac{2x}{(x - 1)}}$

$=> \dfrac{x^2 + 3}{2x (x + 1)}$

Answer 2

$\dfrac{\frac{2}{x-1}+\frac{x-1}{x+1}}{1+\frac{x+1}{x-1}}=\dfrac{1}{2}\\ \\ \dfrac{\frac{2\left(x+1 \right )}{\left(x-1 \right )\left(x+1 \right )}+\frac{\left(x-1 \right )\left(x-1 \right )}{\left(x-1 \right )\left(x+1 \right )}}{\frac{x-1}{x-1}+\frac{x+1}{x-1}}=\dfrac{1}{2}\\ \\ \dfrac{\left[\;\frac{2\left(x+1 \right )+\left(x-1 \right )\left(x-1 \right )}{\left(x-1 \right )\left(x+1 \right )}\; \right ]}{\frac{x-1+x+1}{x-1}}=\dfrac{1}{2}\\ \\ \dfrac{\left[\;\frac{2x+2+x^{2}-x-x+1}{\left(x-1 \right )\left(x+1 \right )}\; \right ]}{\frac{2x}{x-1}}=\dfrac{1}{2}\\ \\ \dfrac{x^{2}+3}{\left(x-1 \right )\left(x+1 \right )}\cdot \dfrac{x-1}{2x}=\dfrac{1}{2}\\ \\ \dfrac{x^{2}+3}{x+1}\cdot \dfrac{1}{2x}=\dfrac{1}{2}\\ \\ \dfrac{x^{2}+3}{\left(x+1 \right )2x}=\dfrac{1}{2}\\ \\ 2\left(x^{2}+3 \right )=1 \cdot \left(x+1 \right )2x\\ \\ 2x^{2}+6=2x^{2}+2x\\ \\ 2x^{2}-2x^{2}+2x=6\\ \\ 2x=6\\ \\ x=\dfrac{6}{2}\\ \\ \boxed{x=3}$