Giải bài tập 6.42 trang 26 Toán 8 tập 2 KNTT

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

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

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

$=\frac{4-4x-6x^{2}+18x^{2}-12}{6x(1-x)}$

$=\frac{12x^{2}-4x-8}{6x(1-x)}$

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

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

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

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

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

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

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

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

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

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

$=\frac{-4x^{2}+8x-2x+4}{(1-x)(2x-1)(2x+1)}$

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

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

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

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

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

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

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

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