Thực hiện phép tính: (x+1) / (x-3) - (1-x) / (x+3)

Giải Toán 8 Bài 6: Phép trừ các phân thức đại số

Video Giải Bài 35 trang 50 Toán 8 Tập 1

Bài 35 trang 50 Toán 8 Tập 1:

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

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

Lời giải

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

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

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

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