Câu 2:
a: Ta có: \(3\sqrt{x}=\sqrt{16x}-5\)
\(\Leftrightarrow3\sqrt{x}-4\sqrt{x}=-5\)
\(\Leftrightarrow-\sqrt{x}=-5\)
hay x=25
b: Ta có: \(\sqrt{4x-8}-\sqrt{9x-18}+4\cdot\sqrt{\dfrac{x-2}{25}}=-3\)
\(\Leftrightarrow2\sqrt{x-2}-3\sqrt{x-2}+\dfrac{4}{5}\sqrt{x-2}=-3\)
\(\Leftrightarrow\sqrt{x-2}=15\)
\(\Leftrightarrow x-2=225\)
hay x=227
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