Bài tập 6.31 trang 13 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

a) Ta có: $\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)$

$=\left[\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right]:\left[\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]$

$=\left[\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right]:\left[\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]$

$=\left[\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right]:\left[\frac{3\left(x-3\right)}{3x\left(x+3\right)}-\frac{x^2}{3x\left(x+3\right)}\right]$

$=\left[\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right]:\left[\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\right]$

$=\frac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}$

$=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}$

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

b) Ta có: $\frac{x+1}{x+2}.\left(\frac{x+2}{x+3}:\frac{x+3}{x+1}\right)$

$=\frac{x+1}{x+2}.\left(\frac{x+2}{x+3}.\frac{x+1}{x+3}\right)$

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

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

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

-- Mod Toán 8 HỌC247