Giải bài 34 trang 50 – SGK Toán lớp 8 tập 1
Dùng quy tắc đổi dấu để thực hiện phép tính:
a) \(\dfrac{4x + 13}{5x(x - 7)} - \dfrac{x - 48}{5x (7 - x)}\)
b) \(\dfrac{1}{x - 5x^2} - \dfrac{25x - 15}{25x^2 - 1}\)
a) \(\dfrac{4x + 13}{5x(x - 7)} - \dfrac{x - 48}{5x (7 - x)}\)
\(= \dfrac{4x + 13}{5x(x - 7)} - \dfrac{x - 48}{-5x (x - 7)}\)
\(= \dfrac{4x + 13}{5x(x - 7)} + \dfrac{x - 48}{5x (x - 7)}\)
\(= \dfrac{4x + 13 + x - 48}{5x(x - 7)}\)
\(= \dfrac{5x - 35}{5x(x - 7)}\)
\(= \dfrac{5(x - 7)}{5x(x - 7)}\)
\(= \dfrac{1}{x}\)
b) \(\dfrac{1}{x - 5x^2} - \dfrac{25x - 15}{25x^2 - 1}\)
\(= \dfrac{-1}{5x^2 - x} - \dfrac{25x - 15}{25x^2 - 1}\)
\(= \dfrac{-1}{x(5x - 1)} - \dfrac{25x - 15}{(5x - 1)(5x + 1)}\)
\(= \dfrac{-(5x + 1)}{x(5x - 1)(5x + 1)} - \dfrac{x(25x - 15)}{x(5x - 1)(5x + 1)}\)
\(= \dfrac{-5x - 1}{x(5x - 1)(5x + 1)} - \dfrac{25x^2 - 15x}{x(5x - 1)(5x + 1)}\)
\(= \dfrac{(-5x - 1) - (25x^2 - 15x)}{x(5x - 1)(5x + 1)} \)
\(= \dfrac{-5x - 1 - 25x^2 + 15x}{x(5x - 1)(5x + 1)} \)
\(= \dfrac{-25x^2 + 10x - 1}{x(5x - 1)(5x + 1)} \)
\(= \dfrac{-(25x^2 - 10x + 1)}{x(5x - 1)(5x + 1)} \)
\(= \dfrac{-(5x - 1)^2}{x(5x - 1)(5x + 1)} \)
\(= \dfrac{-(5x - 1)}{x(5x + 1)}\)
\(= \dfrac{1 - 5x}{x(5x + 1)} \)
Lưu ý: \(\dfrac{-A}{B} = \dfrac{A}{-B}\)