$a)\sqrt[]{3x+10}=4$
Điều kiện xác định : $3x+10≥0⇔x≥\dfrac{-10}{3}$
$⇒3x+10=16$
$⇔3x=6$
$⇔x=2(n)$
Vậy $S=\{2\}$
$b)5\sqrt[]{9x-9}-\sqrt[]{4x-4}-\sqrt[]{x-1}=36$
$⇔5\sqrt[]{3^2(x-1)}-\sqrt[]{2^2(x-1)}-\sqrt[]{x-1}=36$
$⇔15\sqrt[]{x-1}-2\sqrt[]{x-1}-\sqrt[]{x-1}=36$
$⇔12\sqrt[]{x-1}=36$
$⇔\sqrt[]{x-1}=3$
Điều kiện xác định: $x-1≥0⇔x≥1$
$⇒x-1=9$
$⇔x=10(n)$
Vậy $S=\{10\}$
$c)\sqrt[]{x^2-8x+16}=2x+7$
$⇔\sqrt[]{(x-4)^2}=2x+7$
$⇔|x-4|=2x+7$
$⇔$\(\left[ \begin{array}{l}x-4=2x+7\\x-4=-2x-7\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-11(l)\\x=-1(n)\end{array} \right.\)
Vậy $S=\{-1\}$
