Bài $1$.
$a$) $(8+x)(6-x) = 0$
$⇒$ \(\left[ \begin{array}{l}x=-8\\x=6\end{array} \right.\)
Vậy $x$ $∈$ `{-6;8}`.
$b$) $x.(x+3) =0$
$⇒$ \(\left[ \begin{array}{l}x=-3\\x=0\end{array} \right.\)
Vậy $x$ $∈$ `{-3;0}`.
$c$) $4 - (7-x) = x - [13.(-2)-4]$
$⇔ 4 - 7 + x = x - [(-26)-4]$
$⇔ -3 + x = x + 30$
$⇔ x - x = 30 + 3$
$⇔ 0x = 33$ (Vô lý)
Vậy $x ∈ \varnothing$.
$d$) $(x-1)(x^2 - 4)=0$
$⇒$ \(\left[ \begin{array}{l}x=1\\x=2\\x=-2\end{array} \right.\)
Vậy $x ∈$ `{-2;1;2}`.
Bài $2$.
$a$) $4.(x-8) < 0$
$⇒ x-8 < 0$
$⇒ x < 8$
$⇒$ $x$ $∈$ `{7;6;5;4;3}`
$b$) $-3.(x-2)<0$
$⇔ 3.(x-2) > 0$
$⇔ x-2 > 0$
$⇔ x > 2$
$⇒$ $x$ $∈$ `{3;4;5;6;7}`