Giải Bài 4 trang 39 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Tính:

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

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

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

Lời giải chi tiết

a)

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

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

b)

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

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

c)

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

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