Giải Bài 19 trang 41 SGK Toán 8 tập 1 – Chân trời sáng tạo

Đề bài

Thực hiện các phép tính sau:

a) ${\displaystyle \frac{8y}{3x^2}}\cdot {\displaystyle \frac{9x^2}{4y^2}}$

b) ${\displaystyle \frac{3x+x^2}{x^2+x+1}}\cdot {\displaystyle \frac{3x^3-3}{x+3}}$

c) ${\displaystyle \frac{2x^2+4}{x-3}}\cdot {\displaystyle \frac{3x+1}{x-1}}:{\displaystyle \frac{x^2+2}{6-2x}}$

d) ${\displaystyle \frac{2x^2}{3y^3}}:\left(-{\displaystyle \frac{4x^3}{21y^2}}\right)$

e) ${\displaystyle \frac{2x+10}{x^3-64}}:{\displaystyle \frac{{\left(x+5\right)}^2}{2x-8}}$

f) ${\displaystyle \frac{1}{x+y}}\left({\displaystyle \frac{x+y}{xy}}-x-y\right)-{\displaystyle \frac{1}{x^2}}:{\displaystyle \frac{y}{x}}$      

Lời giải chi tiết

a)

${\displaystyle \frac{8y}{3x^2}}\cdot {\displaystyle \frac{9x^2}{4y^2}}$ $={\displaystyle \frac{72x^{2}y}{12x^2{y}^2}}={\displaystyle \frac{6}{y}}$            

b)

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

c)

${\displaystyle \frac{2x^2+4}{x-3}}\cdot {\displaystyle \frac{3x+1}{x-1}}:{\displaystyle \frac{x^2+2}{6-2x}}$ $={\displaystyle \frac{2\left(x^2+2\right)}{x-3}}\cdot {\displaystyle \frac{3x+1}{x-1}}\cdot {\displaystyle \frac{6-2x}{x^2+2}}={\displaystyle \frac{2\left(x^2+2\right)}{x-3}}\cdot {\displaystyle \frac{3x+1}{x-1}}\cdot {\displaystyle \frac{-2\left(x-3\right)}{x^2+2}}={\displaystyle \frac{-4\left(3x+1\right)}{x-1}}$

d)

${\displaystyle \frac{2x^2}{3y^3}}:\left(-{\displaystyle \frac{4x^3}{21y^2}}\right)$ $={\displaystyle \frac{2x^2}{3y^3}}\cdot {\displaystyle \frac{-21y^2}{4x^3}}={\displaystyle \frac{-7}{2xy}}$

e)

${\displaystyle \frac{2x+10}{x^3-64}}:{\displaystyle \frac{{\left(x+5\right)}^2}{2x-8}}$ $={\displaystyle \frac{2x+10}{x^3-64}}\cdot {\displaystyle \frac{2x-8}{{\left(x+5\right)}^2}}={\displaystyle \frac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}}\cdot {\displaystyle \frac{2\left(x-4\right)}{{\left(x+5\right)}^2}}={\displaystyle \frac{4}{\left(x^2+4x+16\right)\left(x+5\right)}}$

f)

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

$\begin{array}{l}={\displaystyle \frac{1}{x+y}}\cdot \left({\displaystyle \frac{x+y}{xy}}-\left(x+y\right)\right)\\ ={\displaystyle \frac{1}{x+y}}\cdot {\displaystyle \frac{x+y}{xy}}-{\displaystyle \frac{1}{x+y}}\cdot \left(x+y\right)\\ ={\displaystyle \frac{1}{xy}}-1\end{array}$