Giải Bài 4 trang 39

Đề bài

Tính:

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

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

c) $\frac{3}{x}-\frac{2}{x}:\frac{1}{x}+\frac{1}{x}\cdot \frac{x^2}{3}$

Lời giải chi tiết

a)

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

$=x\left(x+1\right)-1=x^2+x-1$  

b)

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

$\begin{array}{c}=\left(\frac{1}{x^2}-\frac{x}{x^2}\right)\cdot \frac{x^2}{y}+\frac{x}{y}\\ =\frac{1-x}{x^2}\cdot \frac{x^2}{y}+\frac{x}{y}\\ =\frac{1-x}{y}+\frac{x}{y}\\ =\frac{1}{y}\end{array}$

c)

$\frac{3}{x}-\frac{2}{x}:\frac{1}{x}+\frac{1}{x}\cdot \frac{x^2}{3}$

$\begin{array}{c}=\frac{3}{x}-\frac{2}{x}\cdot \frac{x}{1}+\frac{1}{x}\cdot \frac{x^2}{3}\\ =\frac{3}{x}-2+\frac{x}{3}\\ =\frac{9}{3x}-\frac{6x}{3x}+\frac{x^2}{3x}\\ =\frac{x^2-6x+9}{3x}\end{array}$