Giải bài 29 trang 50 – SGK Toán lớp 8 tập 1

Hướng dẫn:

Muốn trừ phân thức $\dfrac{A}{B}$ cho phân thức $\dfrac{C}{D},$ ta cộng $\dfrac{A}{B}$ với phân thức đối của $\dfrac{C}{D}:$

$\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{A}{B} + \left(-\dfrac{C}{D}\right)$

Bài giải

a) $\dfrac{4x - 1}{3x^2y} - \dfrac{7x - 1}{3x^2y}$

$= \dfrac{4x - 1}{3x^2y} + \dfrac{-(7x - 1)}{3x^2 - 1}$

$= \dfrac{4x - 1}{3x^2y} + \dfrac{-7x + 1}{3x^2 - 1}$

$= \dfrac{(4x - 1) + (-7x + 1)}{3x^2y}$

$= \dfrac{4x - 1 - 7x + 1}{3x^2y}$

$= \dfrac{-3x}{3x^2y}$

$= \dfrac{-1}{xy}$

b) $\dfrac{4x + 5}{2x - 1} - \dfrac{5 - 9x}{2x - 1}$

$= \dfrac{4x + 5}{2x - 1} + \dfrac{-(5 - 9x)}{2x - 1}$

$= \dfrac{4x + 5}{2x - 1} + \dfrac{-5 + 9x}{2x - 1}$

$= \dfrac{(4x + 5) + (-5 + 9x)}{2x - 1}$

$= \dfrac{4x + 5 - 5 + 9x}{2x - 1}$

$= \dfrac{13x}{2x - 1}$

c) $\dfrac{11x}{2x - 3} - \dfrac{x - 18}{3 - 2x}$

$= \dfrac{11x}{2x - 3} + \dfrac{x - 18}{-(3 - 2x)}$

$= \dfrac{11x}{2x - 3} + \dfrac{x - 18}{2x - 3}$

$= \dfrac{11x + x - 18}{2x - 3}$

$= \dfrac{12x - 18}{2x - 3}$

$= \dfrac{6(2x - 3)}{2x - 3}$

$= 6$

d) $\dfrac{2x - 7}{10x - 4} - \dfrac{3x + 5}{4 - 10x}$

$= \dfrac{2x - 7}{10x - 4} + \dfrac{3x + 5}{-(4 - 10x)}$

$= \dfrac{2x - 7}{10x - 4} + \dfrac{3x + 5}{10x - 4}$

$= \dfrac{2x - 7 + 3x + 5}{10x - 4}$

$= \dfrac{5x - 2}{2(5x - 2)}$

$= \dfrac{1}{2}$