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

a) Ta có: $\frac{4y}{3x^2}.\frac{5x^3}{2y^3}=\frac{4y.5x^3}{3x^2.2y^3}=\frac{2.5x}{3.y^2}=\frac{10x}{3y^2}$

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

c) Ta có: $\frac{2x+x^2}{x^2-x+1}.\frac{3x^3+3}{3x+6}=\frac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}$

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

-- Mod Toán 8 HỌC247