Question

Pe 5. Urgent!!/!!!!!!!!

Answer 1

$x^2-1=(x+1)(x-1)\\\displaystyle E(x)=\left(\frac{x}{x+1}-\frac{1}{1-x}-\frac{2x}{x^2-1}\right):\left(x-\frac{4x}{x+1}+1\right)=\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{2x}{x^2-1}\right):\frac{x(x+1)-4x+x+1}{x+1}=\frac{x(x-1)+x+1-2x}{x^2-1}\cdot\frac{x+1}{x^2+x-4x+x+1}=\frac{x^2-x+x+1-2x}{(x-1)(x+1)}\cdot\frac{x+1}{x^2-2x+1}=\frac{x^2-2x+1}{(x-1)(x+1)}\cdot\frac{x+1}{(x-1)^2}=\frac{(x-1)^2}{(x-1)(x+1)}\cdot\frac{x+1}{(x-1)^2}=\frac{1}{x-1}$