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

##### 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)

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