Thực hiện các phép tính sau: (2x+1)/(2x-1) - (2x-1)/(2x+1)

Giải Toán 8 Bài Ôn tập chương 2

Video Giải Bài 58 trang 62 Toán 8 Tập 1

Bài 58 trang 62 Toán 8 Tập 1:

a)$\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5};$

b)$\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right);$

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

Lời giải

a) $\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}$

$\begin{array}{l}=\left.\frac{{\left(2x+1\right)}^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{{\left(2x-1\right)}^2}{\left(2x-1\right)\left(2x+1\right)}\right]:\frac{4x}{5\left(2x-1\right)}\\ =\left.\frac{{\left(2x+1\right)}^2-{\left(2x-1\right)}^2}{\left(2x-1\right)\left(2x+1\right)}\right]:\frac{4x}{5\left(2x-1\right)}\\ =\left.\frac{\left(2x+1-2x+1\right)\left(2x+1+2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}\right]:\frac{4x}{5\left(2x-1\right)}\\ =\left.\frac{2.4x}{\left(2x-1\right)\left(2x+1\right)}\right]:\frac{4x}{5\left(2x-1\right)}\\ =\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5\left(2x-1\right)}{4x}\\ =\frac{8x.5\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right).4x}\\ =\frac{10}{2x+1}.\end{array}$

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

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

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

$\begin{array}{l}\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left(\frac{1}{x^2-2x+1}-\frac{1}{x^2-1}\right)\\ =\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left.\frac{1}{{\left(x-1\right)}^2}-\frac{1}{\left(x-1\right)\left(x+1\right)}\right]\\ =\frac{1}{x-1}-\frac{x^3-x}{x^2+1}.\left.\frac{x+1}{{\left(x-1\right)}^2\left(x+1\right)}-\frac{x-1}{{\left(x-1\right)}^2\left(x+1\right)}\right]\\ =\frac{1}{x-1}-\frac{x\left(x^2-1\right)}{x^2+1}.\left.\frac{x+1-x+1}{{\left(x-1\right)}^2\left(x+1\right)}\right]\\ =\frac{1}{x-1}-\frac{x\left(x-1\right)\left(x+1\right)}{x^2+1}.\frac{2}{{\left(x-1\right)}^2\left(x+1\right)}\\ =\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\\ =\frac{x^2+1}{\left(x-1\right)\left(x^2+1\right)}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\\ =\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x^2+1\right)}\\ =\frac{{\left(x-1\right)}^2}{\left(x-1\right)\left(x^2+1\right)}=\frac{x-1}{x^2+1}.\end{array}$