$a$, $5$ - ($10 -x$) =$7$
⇔ $10 - x $ =$5-7$
⇔ $10 - x $ = $-2$
⇔ $x = 10$ - ($-2$)
⇔ $ x =12$
Vậy $ x =12$
$b$,($4x – 2$)($x + 5$) $= 0$
⇔\(\left[ \begin{array}{l}4x-2=0\\x+5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{1}{2} \\x=-5\end{array} \right.\)
Vậy $x$ ∈ {$\frac{1}{2}$;$-5$}
$c$,$2x – 9 = -8 – 9$
⇔$2x =-8 -9 +9$
⇔ $2x =-8$
⇔ $x = -8 : 2$
⇔ $x= -4$
Vậy $x= -4$
$d$,$3$.|$x-1$| $- 27 =0$
⇔ $3$.|$x-1$| $=27$
⇔ |$x-1$| = $27 : 3$
⇔ |$x-1$| =$9$
⇒\(\left[ \begin{array}{l}x-1=9\\x-1=-9\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=10\\x-1=-8\end{array} \right.\)
Vậy $x$ ∈ {$-8;10$}
$e$,$5.$($3x + 8$) $–7.$($2x + 3$) $= 16$
⇔$15x + 40 -14 x -21 =16$
⇔ $x =16 + 21 - 40$
⇔ $x= -3$
Vậy $x= -3$
