Thực hành 1 trang 37 SGK Toán 8 Chân trời sáng tạo Tập 1 - KNTT

a) Ta có: $\frac{3a^2}{10b^3}.\frac{15b}{9a^4}=\frac{3a^2.15b}{10b^3.9a^4}=\frac{3a^2\mathrm{.3.5}b}{2.5b.b^2.3a^2.3a^2}=\frac{1}{2a^2{b}^2}$

b) Ta có: $\frac{x-3}{x^2}.\frac{4x}{x^2-9}=\frac{\left(x-3\right).4x}{x.x.\left(x+3\right)\left(x-3\right)}=\frac{4}{x\left(x+3\right)}$

c) Ta có: $\frac{a^2-6a+9}{a^2+3a}.\frac{2a+6}{a-3}=\frac{{\left(a-3\right)}^2.2\left(a+3\right)}{a\left(a+3\right).\left(a-3\right)}=\frac{2\left(a-3\right)}{a}$

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

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

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

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

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

Cách khác: Sử dụng tính chất phân phối của phép nhân đối với phép cộng

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

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

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

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

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

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

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

-- Mod Toán 8 HỌC247