a) Ta có: \(\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12\)
\(\Leftrightarrow\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12\)
\(\Leftrightarrow3x=1\)
hay \(x=\dfrac{1}{3}\)
b) Ta có: \(\sqrt{4x-4}+\dfrac{1}{9}\sqrt{9x-9}=12\)
\(\Leftrightarrow2\sqrt{x-1}+\dfrac{1}{3}\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=\dfrac{1296}{49}\)
hay \(x=\dfrac{1345}{49}\)






