Đáp án:
$C1 \dfrac{-2x+1}{-3} =\dfrac{3x-5}{-4}$
$⇔\dfrac{-4(-2x+1)}{-3.(-4)} = \dfrac{-3(3x-5)}{-4.(-3)}$
$⇔\dfrac{-4(-2x+1)}{12} = \dfrac{-3(3x-5)}{12}$
$⇔-4(-2x+1) = -3(3x-5)$
$⇔ 8x -4 = -9x +15$
$⇔8x +9x = 15 +4$
$⇔17x=19$
$⇔x = \dfrac{19}{17}$
Vậy $x=\dfrac{19}{17}$
$C2 \dfrac{x-3}{-7} = \dfrac{-3x+2}{5}$
$⇔\dfrac{5(x-3)}{-7 . 5} = \dfrac{-7(-3x+2)}{5 . (-7)}$
$⇔\dfrac{5(x-3)}{-35} = \dfrac{-7(-3x+2)}{-35}$
$⇔ 5(x-3) = -7(-3x+2)$
$⇔5x-15 = 21x -14$
$⇔5x-21x = -14 +15$
$⇔-16x = 1$
$⇔x = -\dfrac{1}{16}$
Vậy $x=-\dfrac{1}{16}$
