Bài 4 trang 39 SGK Toán 8 Chân trời sáng tạo Tập 1 - KNTT

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

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

$=\frac{-\left(x-1\right)+x\left(x+1\right)\left(x-1\right)}{x}.\frac{x}{x-1}$

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

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

= x² + 1 – 1;

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

$=\frac{1}{x^2}.\frac{x^2}{y}-\frac{1}{x}.\frac{x^2}{y}+\frac{x}{y}$

$=\frac{1}{y}-\frac{x}{y}+\frac{x}{y}$

$=\frac{1}{y};$

c) Ta có: $\frac{3}{x}-\frac{2}{x}:\frac{1}{x}+\frac{1}{x}.\frac{x^2}{3};$

$=\frac{3}{x}-\frac{2}{x}.\frac{x}{1}+\frac{x}{3}$

$=\frac{3}{x}-2+\frac{x}{3}$

$=\frac{3^2-2.x.3+x^2}{3x}$

$=\frac{9-6x+x^2}{3x}$

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

-- Mod Toán 8 HỌC247