a: Ta có: \(\sqrt{4-3x}=8\)
\(\Leftrightarrow4-3x=64\)
\(\Leftrightarrow3x=-60\)
hay x=-20
b: ta có: \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1\)
\(\Leftrightarrow2\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)
\(\Leftrightarrow x-2=\dfrac{1}{4}\)
hay \(x=\dfrac{9}{4}\)
\(\left\{{}\begin{matrix}8>0\left(luondung\right)\\4-3x=64\end{matrix}\right.\) \(\Leftrightarrow x=-20\left(ktm\right)\)