a, $|x-3|=4$
$⇔$\(\left[ \begin{array}{l}x-3=4\\x-3=-4\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=7\\x=-1\end{array} \right.\)
Vậy $S=\{7;-1\}$
b, $|x-3|=|x+4|$
$⇔$\(\left[ \begin{array}{l}x-3=x+4\\x-3=-x-4\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x-x=4+3\\x+x=-4+3\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}0x=7(vô.lý)\\2x=-1⇔x=-1/2\end{array} \right.\)
Vậy $S=\{-1/2\}$
c, $|x+3|+|x^2-9|=0$
Vì: $|x+3≥0;|x^2-9|≥0$
$⇒|x+3|+|x^2-9|≥0$
Dấu "=" xảy ra khi $\left \{ {{x+3=0} \atop {x^2-9=0}} \right.$ $⇔\left \{ {{x=-3} \atop {x=±3}} \right.$
Vậy $S=\{-3\}$ (chọn nghiệm chung)
