Thực hành 2 trang 34 SGK Toán 8 Chân trời sáng tạo Tập 1 - KNTT

a) Ta có: $\frac{a}{a-3}-\frac{3}{a+3}$

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

$=\frac{a^2+3a-\left(3a-9\right)}{\left(a-3\right)\left(a+3\right)}$

$=\frac{a^2+3a-3a+9}{\left(a-3\right)\left(a+3\right)}$

$=\frac{a^2+9}{\left(a-3\right)\left(a+3\right)}$;

b) Ta có: $\frac{1}{2x}+\frac{2}{x^2}=\frac{x}{2x^2}+\frac{4}{2x^2}=\frac{x+4}{2x^2}$;

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

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

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

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

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

-- Mod Toán 8 HỌC247