Bài tập 6.30 trang 12 SBT Toán 8 Tập 2 Kết nối tri thức - KNTT

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

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

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

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

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

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

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

b) Ta có: $\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}$

$=\left[\frac{3x\left(3x+1\right)}{\left(1-3x\right)\left(3x+1\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right]:\frac{2x\left(3x+5\right)}{{\left(1-3x\right)}^2}$

$=\left[\frac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right]:\frac{2x\left(3x+5\right)}{{\left(1-3x\right)}^2}$

$=\left[\frac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\right]:\frac{2x\left(3x+5\right)}{{\left(1-3x\right)}^2}$

$=\left[\frac{3x^2+5x}{\left(1-3x\right)\left(3x+1\right)}\right]:\frac{2x\left(3x+5\right)}{{\left(1-3x\right)}^2}$

$=\frac{x\left(3x+5\right)}{\left(1-3x\right)\left(3x+1\right)}:\frac{2x\left(3x+5\right)}{{\left(1-3x\right)}^2}$

$=\frac{x\left(3x+5\right)}{\left(1-3x\right)\left(3x+1\right)}.\frac{{\left(1-3x\right)}^2}{2x\left(3x+5\right)}$

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

-- Mod Toán 8 HỌC247