a, 2-x/4=2(x+1)/5-3(2x-5)/10 b, (3x+6)(x-7)=0 Giải giúp m vs ạ

$a) \dfrac{2-x}{4}=\dfrac{2(x+1)}{5}-\dfrac{3(2x-5)}{10}$

$\to \dfrac{5(2-x)}{20}=\dfrac{4(x+1)}{20}-\dfrac{6(2x-5)}{20}$

$\to 10-5x=4x+4-12x+30$

$\to -5x-4x+12x=4+30-10$

$\to 3x=24$

$\to x=8$

Vậy $S=\{8\}$

$b) (3x+6)(x-7)=0$

\(\to\left[ \begin{array}{l}3x+6=0\\x-7=0\end{array} \right.\)

\(\to\left[ \begin{array}{l}3x=-6\\x=7(tm)\end{array} \right.\)

\(\to\left[ \begin{array}{l}x=-2(tm)\\x-7=0(tm)\end{array} \right.\)

Vậy $S=\{-2;7\}$