Giải bài 6.42 trang 26

Đề bài

Rút gọn biểu thức sau:

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

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 chi tiết

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

c) Ta có: 

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

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

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

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

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

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

Do đó

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

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

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

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

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

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

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

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

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

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

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