Giải bài 4 trang 16 vở thực hành Toán 8 tập 2

Đề bài

Thực hiện các phép tính sau:

a) $\frac{x^2+4x+4}{x^2-4}+\frac{x}{2-x}+\frac{4-x}{5x-10}$

b) $\frac{x}{x^2+1}-\left(\frac{3}{x+6}+\frac{x-2}{x+4}\right)+\left[\frac{3}{x+6}-\left(\frac{1}{x^2+1}-\frac{x-2}{x+4}\right)\right]$

Lời giải chi tiết

a)

$\begin{array}{c}\frac{x^2+4x+4}{x^2-4}+\frac{x}{2-x}+\frac{4-x}{5x-10}\\ =\frac{{\left(x+2\right)}^2}{\left(x+2\right)\left(x-2\right)}+\frac{-x}{x-2}+\frac{4-x}{5\left(x-2\right)}\\ =\frac{x+2}{x-2}+\frac{-x}{x-2}+\frac{4-x}{5\left(x-2\right)}\\ =\frac{2}{x-2}+\frac{4-x}{5(x-2)}=\frac{-x+14}{5\left(x-2\right)}\end{array}$

b)

$\begin{array}{c}\frac{x}{x^2+1}-\left(\frac{3}{x+6}+\frac{x-2}{x+4}\right)+\left[\frac{3}{x+6}-\left(\frac{1}{x^2+1}-\frac{x-2}{x+4}\right)\right]\\ =\frac{x}{x^2+1}-\frac{3}{x+6}-\frac{x-2}{x+4}+\frac{3}{x+6}-\frac{1}{x^2+1}+\frac{x-2}{x+4}\\ \begin{array}{c}=\left(\frac{x}{x^2+1}-\frac{1}{x^2+1}\right)+\left(\frac{3}{x+6}-\frac{3}{x+6}\right)+\left(\frac{x-2}{x+4}-\frac{x-2}{x+4}\right)\\ =\frac{x-1}{x^2+1}+0+0=\frac{x-1}{x^2+1}\end{array}\end{array}$