$a)x-7\sqrt[]{x}-8=0$ với $(x≥0)$
$⇔(x+\sqrt[]{x})-(8\sqrt[]{x}+8)=0$
$⇔\sqrt[]{x}(x+1)-8(\sqrt[]{x}+1)=0$
$⇔(\sqrt[]{x}+1)(\sqrt[]{x}-8)=0$
$⇔$\(\left[ \begin{array}{l}\sqrt[]{x}+1=0\\\sqrt[]{x}-8=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}\sqrt[]{x}=-1(l)\\\sqrt[]{x}=8\end{array} \right.\)
$⇔x=64$
Vậy $x=64$
$b)\sqrt[]{9x^2}=4$
$⇔9x^2=4^2$
$⇔9x^2=36$
$⇔x^2=4$
$⇔$\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy $x=±2$
