Thực hiện các phép tính:

a) $\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2};$                    b) $\frac{3x+1}{{\left(x-1\right)}^2}-\frac{1}{x+1}+\frac{x+3}{1-x^2}$.

Giải

a) $\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}=\frac{x+1}{x-3}+\frac{-\left(1-x\right)}{x+3}+\frac{2x\left(1-x\right)}{-\left(9-x^2\right)}$

$$\begin{gathered} =\frac{x+1}{x-3}+\frac{x-1}{x+3}+\frac{2x\left(1-x\right)}{x^2-9}=\frac{x+1}{\left(x-3\right)}+\frac{x-1}{x+3}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)} \\ =\frac{\left(x+1\right)\left(x+3\right)+\left(x-1\right)\left(x-3\right)+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\frac{x^2+4x+3+x^2-4x+3+2x-2x^2}{\left(x-3\right)\left(x+3\right)} \\ =\frac{2x+6}{\left(x-3\right)\left(x+3\right)}=\frac{2(x+3)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{x-3} \end{gathered}$$

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

$$\begin{gathered} =\frac{3x+1}{{\left(x-1\right)}^2}+\frac{-1}{x+1}+\frac{-\left(x+3\right)}{x^2-1}=\frac{3x+1}{{\left(x-1\right)}^2}+\frac{-1}{x+1}+\frac{-\left(x+3\right)}{\left(x-1\right)\left(x+1\right)} \\ =\frac{\left(3x+1\right)\left(x+1\right)-{\left(x-1\right)}^2-\left(x+3\right)\left(x-1\right)}{{\left(x-1\right)}^2\left(x+1\right)}=\frac{3x^2+4x+1-(x^2-2x+1)-(x^2+2x-3)}{{\left(x-1\right)}^2\left(x+1\right)} \\ =\frac{3x^2+4x+1-x^2+2x-1-x^2-2x+3)}{{\left(x-1\right)}^2\left(x+1\right)}=\frac{x^2+4x+3}{{\left(x-1\right)}^2\left(x+1\right)}=\frac{x^2+x+3x+3}{{\left(x-1\right)}^2\left(x+1\right)} \\ =\frac{x\left(x+1\right)+3\left(x+1\right)}{{\left(x-1\right)}^2\left(x+1\right)}=\frac{\left(x+3\right)\left(x+1\right)}{{\left(x-1\right)}^2\left(x+1\right)}=\frac{x+3}{{\left(x-1\right)}^2} \end{gathered}$$