Bài 3 trang 39 SGK Toán 8 Chân trời sáng tạo Tập 1 - KNTT

a) Ta có: $\frac{4x^2+2}{x-2}.\frac{3x+2}{x-4}.\frac{4-2x}{2x^2+1}$

$=\frac{2\left(2x^2+1\right).\left(3x+2\right).2\left(2-x\right)}{\left(x-2\right).\left(x-4\right).\left(2x^2+1\right)}$

$=\frac{2.\left(3x+2\right)\mathrm{.2.}\left(-1\right)\left(x-2\right)}{\left(x-2\right).\left(x-4\right)}$

$=\frac{2\left(3x+2\right)\mathrm{.2.}\left(-1\right)}{x-4}$

$=\frac{-4\left(3x+2\right)}{x-4}$;

b) Ta có: $\frac{x+3}{x}.\frac{x+2}{x^2+6x+9}:\frac{x^2-4}{x^2+3x}$.

$=\frac{\left(x+3\right).\left(x+2\right)}{x.{\left(x+3\right)}^2}.\frac{x^2+3x}{x^2-4}$

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

$=\frac{1}{x-2}$.

-- Mod Toán 8 HỌC247