Đáp án:
$a,S=\{-4;10\}$
$b,S=\{11\}$
Giải thích các bước giải:
a) ĐKXĐ: $x\in\mathbb R$
$\sqrt{x^2-6x+9}=7$
$⇔\sqrt{(x-3)^2}=7$
$⇔|x-3|=7$
$⇔\left[ \begin{array}{l}x-3=7\\x-3=-7\end{array} \right.⇔\left[ \begin{array}{l}x=10\\x=-4\end{array} \right.$
Vậy $S=\{-4;10\}$
b) ĐKXĐ: $x\ge -\dfrac{3}{2}$
$4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15$
$⇔4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$⇔3\sqrt{2x+3}=15$
$⇔\sqrt{2x+3}=5$
$⇔2x+3=25$
$⇔2x=22$
$⇔x=11\,(TM)$
Vậy $S=\{11\}$.
