`C1` `\frac{x-3}{4}` = `\frac{2-x}{-7}` `C2` `\frac{2x-1}{-3}` = `\frac{4-5x}{-7}` `C3` `\frac{x+8}{-21}` = `\frac{-3x+2}{13}`

Đáp án:

 $C1 \dfrac{x-3}{4}=\dfrac{2-x}{-7}$

$⇔\dfrac{-7(x-3)}{-28} = \dfrac{4(2-x)}{-28}$

$⇔-7(x-3) = 4(2-x)$

$⇔-7x+21 = 8 -4x$

$⇔-7x+4x=8-21$

$⇔-3x =-13$

$⇔x=-13 : (-3)$

$⇔x = \dfrac{13}{3}$

Vậy $x=\dfrac{13}{3}$

$C2 \dfrac{2x-1}{-3} = \dfrac{4-5x}{-7}$

$⇔\dfrac{-7(2x-1)}{21} = \dfrac{-3(4-5x)}{21}$

$⇔-7(2x-1) = -3(4-5x)$

$⇔-14x +7 = -12 +15x$

$⇔-14x-15x=-12-7$

$⇔-29x = -19$

$⇔x = -19 : (-29)$

$⇔x=\dfrac{19}{29}$

Vậy $x=\dfrac{19}{29}$

$C3 \dfrac{x+8}{-21} =\dfrac{-3x+2}{13}$

$⇔\dfrac{13(x+8)}{-273} =\dfrac{-21(-3x+2)}{-273}$

$⇔ 13(x+8) = -21(-3x+2)$

$⇔13x+104=63x -42$

$⇔ 13x-63x = -42 -104$

$⇔-50x =-146$

$⇔x = -146 : (-50)$

$⇔x=\dfrac{73}{25}$

Vậy $x=\dfrac{73}{25}$