Bài 18 trang 41 Toán 8 Tập 1 Chân trời sáng tạo

Giải Toán 8 Bài tập cuối chương 1 - Chân trời sáng tạo

Bài 18 trang 41 Toán 8 Tập 1: Thực hiện các phép tính sau:

a) $\frac{2x^2-1}{x-2}+\frac{-x^2-3}{x-2}$;

b) $\frac{x}{x+y}+\frac{y}{x-y}$;

c) $\frac{1}{x-1}-\frac{2}{x^2-1}$;

d) $\frac{x+2}{x^2+xy}-\frac{y-2}{xy+y^2}$;

e) $\frac{1}{2x^2-3x}-\frac{1}{4x^2-9}$;

g) $\frac{2x}{9-x^2}+\frac{1}{x-3}-\frac{1}{x+3}$;

Lời giải:

a) $\frac{2x^2-1}{x-2}+\frac{-x^2-3}{x-2}$

$=\frac{2x^2-1-x^2-3}{x-2}=\frac{x^2-4}{x-2}$

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

b) $\frac{x}{x+y}+\frac{y}{x-y}$

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

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

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

c) $\frac{1}{x-1}-\frac{2}{x^2-1}$

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

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

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

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

$=\frac{1}{x+1}$.

d) $\frac{x+2}{x^2+xy}-\frac{y-2}{xy+y^2}$

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

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

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

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

$=\frac{2y+2x}{xy\left(x+y\right)}=\frac{2\left(x+y\right)}{xy\left(x+y\right)}=\frac{2}{xy}$

e) $\frac{1}{2x^2-3x}-\frac{1}{4x^2-9}$

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

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

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

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

g) $\frac{2x}{9-x^2}+\frac{1}{x-3}-\frac{1}{x+3}$

$=\frac{-2x}{x^2-9}+\frac{1}{x-3}-\frac{1}{x+3}$

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

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

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

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

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

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

Lời giải bài tập Toán 8 Bài tập cuối chương 1 hay, chi tiết khác: