Bài 6.42 trang 26 Toán 8 Tập 2 | Kết nối tri thức Giải Toán lớp 8

Giải Toán 8 Bài tập cuối chương 6 (trang 26)

Bài 6.42 trang 26 Toán 8 Tập 2: Rút gọn các biểu thức sau:

a) $\frac{2}{3x}+\frac{x}{x-1}+\frac{6x^2-4}{2x(1-x)}$ ;

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

c) $\left(\frac{2}{x+2}-\frac{2}{1-x}\right).\frac{x^2-4}{4x^2-1}$ ;

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

Lời giải:

a) $\frac{2}{3x}+\frac{x}{x-1}+\frac{6x^2-4}{2x(1-x)}$

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

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

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

$=\frac{2-2x-3x^2+9x^2-6}{3x(1-x)}$

$=\frac{6x^2-2x-4}{3x(1-x)}$.

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

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

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

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

$=\frac{-x}{1-x^3}=\frac{x}{x^3-1}$.

c) $\left(\frac{2}{x+2}-\frac{2}{1-x}\right).\frac{x^2-4}{4x^2-1}$

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

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

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

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

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

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

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

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

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

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