Giải bài 58 trang 62 - Sách giáo khoa Toán 8 tập 1

Đề bài

 Thực hiện các phép tính sau:

a) \((\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}):\dfrac{4x}{10x-5}\)

b) \((\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}):(\dfrac{1}{x}+x-2)\)

c) \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2})\)

Hướng dẫn giải

a) \((\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}):\dfrac{4x}{10x-5}\)\(=\dfrac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\dfrac{10x-5}{4x}\)

\(=\dfrac{(2x+1-2x+1)(2x+1+2x-1)}{(2x-1)(2x+1)}\) \(.\dfrac{5(2x-1)}{4x}\)

\(=\dfrac{2.4x.5(2x-1)}{(2x-1)(2x+1).4x}=\dfrac{10}{2x+1}\)

b) \((\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}):(\dfrac{1}{x}+x-2)\) \(=(\dfrac{1}{x(x+1)}-\dfrac{2-x}{x+1}):(\dfrac{1}{x}+x-2)\)

\(=\dfrac{1-x(2-x)}{x(x+1)}:\dfrac{1+x^2-2x}{x}\)\(=\dfrac{x^2-2x+1}{x(x+1)}.\dfrac{x}{x^2-2x+1}=\dfrac{1}{x+1}\)

(ĐK : x \(\neq\) 0 ; x \(\neq\) 1 và x \(\neq\) -1)

c) \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2})\)

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

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

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

(ĐK:  x \(\neq\) \(\pm\)1)

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