$a) |2x-3| - |5| = |-11| - 3$
$⇔ |2x-3| - 5 = 11 - 3$
$⇔ |2x-3| = 13$
$⇒$ \(\left[ \begin{array}{l}2x-3=13\\2x-3=-13\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=8\\x=-5\end{array} \right.\)
Vậy $x$ $∈$ `{8;-5}`
$b) | x | + | y | = 0$
Vì : $|x|;|y|$ $≥$ $0$ $∀$ $x$
$⇒$ $\left \{ {{x=0} \atop {y=0}} \right.$
Vậy `(x;y)=(0;0)`
$c) | x - 3 | + | y + 2 | = 0$
Vì : $|x-3|;|y+2|$ $≥$ $0$ $∀$ $x$
$⇒$ $\left \{ {{x-3=0} \atop {y+2=0}} \right.$
$⇔$ $\left \{ {{x=3} \atop {y=-2}} \right.$
Vậy `(x;y)=(3;-2)`
$d) x.(x + 7) = 0$
$⇒$ \(\left[ \begin{array}{l}x=0\\x=-7\end{array} \right.\)
Vậy $x$ $∈$ `{0;-7}`
$e)(x + 12).(x-3) = 0 $
$⇒$ \(\left[ \begin{array}{l}x=-12\\x=3\end{array} \right.\)
Vậy $x$ $∈$ `{-12;3}`
$f)(-x + 5).(3 – x ) = 0$
$⇒$ \(\left[ \begin{array}{l}x=5\\x=3\end{array} \right.\)
Vậy $x$ $∈$ `{5;3}`
$g) x.(2 + x).( 7 – x) = 0 $
$⇒$ \(\left[ \begin{array}{l}x=0\\x=-2\\x=7\end{array} \right.\)
Vậy $x$ $∈$ `{0;-2;7}`
$ h) (x - 1).(x +2).(-x -3) = 0$
$⇒$ \(\left[ \begin{array}{l}x=1\\x=-2\\x=-3\end{array} \right.\)
Vậy $x$ $∈$ `{1;-2;-3}`
