Bài 2 trang 35 SGK Toán 8 Chân trời sáng tạo Tập 1 - KNTT

a) Ta có: $\frac{1}{2a}+\frac{2}{3b}=\frac{3b}{2a.3b}+\frac{2.2a}{3b.2a}=\frac{3b+4a}{6ab}$

b) Ta có: $\frac{x-1}{x+1}-\frac{x+1}{x-1}$

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

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

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

$=\frac{-4x}{x^2-1}$;

c) Ta có: $\frac{x+y}{xy}-\frac{y+z}{yz}$

$=\frac{\left(x+y\right)z}{xyz}-\frac{\left(y+z\right)x}{xyz}$

$=\frac{xz+yz-\left(xy+xz\right)}{xyz}$

$=\frac{xz+yz-xy-xz}{xyz}$

$=\frac{yz-xy}{xyz}=\frac{y\left(z-x\right)}{xyz}=\frac{z-x}{xz}$

d) Ta có: $\frac{2}{x-3}-\frac{12}{x^2-9}$

$=\frac{2}{x-3}-\frac{12}{\left(x-3\right)\left(x+3\right)}$

$=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{12}{\left(x-3\right)\left(x+3\right)}$

$=\frac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\frac{2x-6}{\left(x-3\right)\left(x+3\right)}$

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

e) $\frac{1}{x-2}+\frac{2}{x^2-4x+4}$

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

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

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

-- Mod Toán 8 HỌC247