Thực hiện các phép tính sau:

a) $\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5};$                                     b) $\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)$

c) $\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)$.

Giải

a) $\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{{\left(2x+1\right)}^2-{\left(2x-1\right)}^2}{\left(2x+1\right)\left(2x-1\right)}.\frac{10x-5}{4x}$

$=\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x+1\right)\left(2x-1\right)}.\frac{5(2x-1)}{4x}=\frac{8x.5(2x-1)}{\left(2x+1\right)\left(2x-1\right).4x}=\frac{10}{2x+1}$

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

c) $\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left(\frac{1}{x^2-2x+1}+\frac{1}{1-x^2}\right)=\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left[\frac{1}{{\left(x-1\right)}^2}-\frac{1}{\left(x-1\right)(x+1)}\right]$

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