Giải bài 58 trang 62 – SGK Toán lớp 8 tập 1
Thực hiện các phép tính sau:
\(a) \,\,\,\, \left(\dfrac{2x + 1}{2x - 1} - \dfrac{2x - 1}{2x + 1}\right) : \dfrac{4x}{10x - 5}; \\ b) \,\,\,\, \left(\dfrac{1}{x^2 + x} - \dfrac{2 - x}{x + 1}\right) : \left(\dfrac{1}{x} + x - 2 \right); \\ c) \,\,\,\, \dfrac{1}{x - 1} - \dfrac{x^3 - x}{x^2 + 1} .\left(\dfrac{1}{x^2 -2x + 1} - \dfrac{1}{1 - x^2}\right).\)
\(a) \,\,\,\, \left(\dfrac{2x + 1}{2x - 1} - \dfrac{2x - 1}{2x + 1}\right) : \dfrac{4x}{10x - 5} \\ = \left(\dfrac{(2x + 1)^2}{(2x - 1)(2x + 1)} - \dfrac{(2x - 1)^2}{(2x - 1)(2x + 1)}\right) : \dfrac{4x}{10x - 5} \\ = \left(\dfrac{(2x + 1)^2 - (2x - 1)^2}{(2x - 1)(2x + 1)} \right) : \dfrac{4x}{10x - 5} \\ = \left(\dfrac{(2x + 1 + 2x - 1)(2x + 1 - 2x + 1)}{(2x - 1)(2x + 1)} \right) . \dfrac{10x - 5}{4x} \\ = \dfrac{4x.2}{(2x - 1)(2x + 1)} . \dfrac{5(2x - 1)}{4x} \\ = \dfrac{10}{2x + 1} \\ b)\,\,\,\, \left(\dfrac{1}{x^2 + x} - \dfrac{2 - x}{x + 1}\right) : \left(\dfrac{1}{x} + x - 2 \right) \\ = \left(\dfrac{1}{x(x + 1)} - \dfrac{2 - x}{x + 1}\right) : \left(\dfrac{1}{x} + \dfrac{x^2}{x} - \dfrac{2x}{x} \right) \\ = \left(\dfrac{1}{x(x + 1)} - \dfrac{x(2 - x)}{x(x + 1)}\right) : \left(\dfrac{1 + x^2 - 2x}{x} \right) \\ = \left(\dfrac{1}{x(x + 1)} - \dfrac{2x - x^2}{x(x + 1)}\right) : \left(\dfrac{x^2 - 2x + 1}{x} \right) \\ = \dfrac{1 - (2x - x^2)}{x(x + 1)} : \dfrac{(x - 1)^2}{x}\\ = \dfrac{1 - 2x + x^2}{x(x + 1)}.\dfrac{x}{(x - 1)^2}\\ = \dfrac{(x - 1)^2}{x(x + 1)}.\dfrac{x}{(x - 1)^2}\\ = \dfrac{1}{x + 1} \\ c) \,\,\,\, \dfrac{1}{x - 1} - \dfrac{x^3 - x}{x^2 + 1} .\left(\dfrac{1}{x^2 -2x + 1} - \dfrac{1}{1 - x^2}\right) \\ = \dfrac{1}{x - 1} - \dfrac{x(x^2 - 1)}{x^2 + 1} .\left(\dfrac{1}{(x - 1)^2} + \dfrac{1}{x^2 - 1}\right) \\ = \dfrac{1}{x - 1} - \dfrac{x(x^2 - 1)}{x^2 + 1} .\left(\dfrac{1}{(x - 1)^2} + \dfrac{1}{(x - 1)(x + 1)}\right) \\ = \dfrac{1}{x - 1} - \dfrac{x(x^2 - 1)}{x^2 + 1} .\left(\dfrac{x + 1}{(x - 1)^2(x + 1)} + \dfrac{x - 1}{(x - 1)^2(x + 1)}\right) \\ = \dfrac{1}{x - 1} - \dfrac{x(x^2 - 1)}{x^2 + 1} .\left(\dfrac{x + 1 + x - 1}{(x - 1)^2(x + 1)} \right) \\ = \dfrac{1}{x - 1} - \dfrac{x(x - 1)(x + 1)}{x^2 + 1} .\dfrac{2x}{(x - 1)^2(x + 1)} \\ = \dfrac{1}{x - 1} - \dfrac{2x^2}{(x^2 + 1)(x - 1)} \\ = \dfrac{x^2 + 1}{(x^2 + 1)(x - 1)} - \dfrac{2x^2}{(x^2 + 1)(x - 1)} \\ = \dfrac{x^2 + 1 - 2x^2}{(x^2 + 1)(x - 1)} \\ = \dfrac{-x^2 + 1 }{(x^2 + 1)(x - 1)} \\ = \dfrac{-(x^2 - 1)}{(x^2 + 1)(x - 1)} \\ = \dfrac{-(x - 1)(x + 1)}{(x^2 + 1)(x - 1)} \\ = \dfrac{-(x + 1)}{x^2 + 1} \)
Lưu ý: \(- \dfrac{1}{1 - x^2} = - \dfrac{1}{-(x^2 - 1)} = \dfrac{1}{x^2 - 1}\)
