Đáp án:
Giải thích các bước giải:
a, $\frac{7x-3}{x-1}= \frac{2}{3}$
$⇔ 3( 7x - 3) = 2(x-1)$
$⇔ 21x -9 = 2x - 2$
$⇔ 21x - 2x= -2 + 9$
$⇔ 19x = 7$
$⇔ x = 7/19$
b, $\frac{5x-1}{3x+2}= \frac{5x-7}{3x-1}$
$⇔ (5x-1)( 3x-1) = (5x-7) ( 3x+2) $
$⇔ 15x^2 - 5x - 3x + 1 = 15x^2 + 10x - 21x - 14 $
$⇔ -8x + 1 = -11x - 14 $
$⇔ -8x + 11x = -14 - 1 $
$⇔ 3x = -15 $
$⇔ x = -5 $
c, $\frac{x+5}{x-5}- \frac{x-5}{x+5}= \frac{20}{x^2-25}$
$⇔ \frac{(x+5)^2-(x-5)^2}{(x+5)(x-5)}= \frac{20}{x^2-25}$
$⇔ \frac{(x+5+x-5)(x+5-x+5)}{(x^2-5^2}= \frac{20}{x^2-25}$
$⇔ \frac{20x}{x^2-25}= \frac{20}{x^2-25}$
$⇔ 20x = 20 $
$⇔ x = 1 $
d, $\frac{x}{2(x-3)}+ \frac{x}{2(x+1)}= \frac{2x}{(x+1)(x-3)}$
$⇔ \frac{2x(x+1)+2x(x-3)}{4(x+1)(x-3)}= \frac{2x}{(x+1)(x-3)}$
$⇔ \frac{2x(x+1+x-3)}{4(x+1)(x-3)}= \frac{2x}{(x+1)(x-3)}$
$⇔ \frac{x(2x-2)}{2(x+1)(x-3)}= \frac{2x}{(x+1)(x-3)}$
$⇔ \frac{2x(x-1)}{2(x+1)(x-3)}= \frac{2x}{(x+1)(x-3)}$
$⇔ \frac{x(x-1)}{(x+1)(x-3)}= \frac{2x}{(x+1)(x-3)}$
$⇔ x(x-1) = 2x $
$⇔ x^2 - x = 2x $
$⇔ x^2 = 3x $
$⇔ x( x - 3 ) = 0 $
$⇔ x = 0$ hoặc $x = 3 $
