Giải bài 2 trang 25 vở thực hành Toán 8 tập 2

Đề bài

Rút gọn biểu thức sau:

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

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

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

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}{3\mathrm{x}}+\frac{x}{x-1}+\frac{6x^2-4}{2x\left(1-x\right)}$$=\frac{2}{3\mathrm{x}}+\frac{-x}{1-x}+\frac{3{\mathrm{x}}^2-2}{x\left(1-x\right)}$$=\frac{2-2x-3x^2+9x^2-6}{3x\left(1-x\right)}$

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

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

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

Do đó

 $\begin{array}{l}\left(\frac{2}{x+2}-\frac{2}{1-x}\right).\frac{x^2-4}{4{\mathrm{x}}^2-1}=\frac{2\left(2x+1\right)}{\left(x+2\right)\left(x-1\right)}.\frac{\left(x-2\right)\left(x+2\right)}{\left(2x-1\right)\left(2x+1\right)}\\ =\frac{2(x-2)}{\left(2x-1\right)\left(x-1\right)}\end{array}$

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

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

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