Thực hiện các phép tính: (x/ (x+1) +1):(1- 3x^2/ (1-x^2))

Giải Toán 8 Bài 9: Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

Video Giải Bài 50 trang 58 Toán 8 Tập 1

Bài 50 trang 58 Toán 8 Tập 1:

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

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

Lời giải

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

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

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

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