Thực hiện các phép tính:

a) $\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x^2}\right);$                     b) $\left(x^2-1\right)\left(\frac{1}{x-1}-\frac{1}{x+1}-1\right)$.

Giải

a) $\left(\frac{x}{x+1}+1\right):\left(1-\frac{3x^2}{1-x^2}\right)=\frac{x+x+1}{x+1}:\frac{1-x^2-3x^2}{1-x^2}=\frac{2x+1}{x+1}:\frac{1-4x^2}{1-x^2}$

$=\frac{2x+1}{x+1}.\frac{1-x^2}{1-4x^2}=\frac{2x+1}{x+1}.\frac{\left(1-x\right)\left(1+x\right)}{\left(1-2x\right)\left(1+2x\right)}=\frac{1-x}{1-2x}.$

b) $\left(x^2-1\right)\left(\frac{1}{x-1}-\frac{1}{x+1}-1\right)=\left(x^2-1\right)\left[\frac{x+1-\left(x-1\right)-(x-1)(x+1)}{(x-1)(x+1)}\right]$

$$\begin{gathered} =\left(x^2-1\right).\frac{x+1-x+1-x^2+1}{(x-1)(x+1)}=\left(x^2-1\right).\frac{3-x^2}{(x-1)(x+1)} \\ =\frac{(x-1)(x+1)\left(3-x^2\right)}{(x-1)(x+1)}=3-x^2. \end{gathered}$$