Ta có: \(\sqrt{3x}+5\sqrt{27x}-16=\sqrt{432x}\)
\(\Leftrightarrow\sqrt{3x}+15\sqrt{3x}-12\sqrt{3x}=16\)
\(\Leftrightarrow\sqrt{3x}=4\)
\(\Leftrightarrow3x=16\)
hay \(x=\dfrac{16}{3}\)
b. \(\sqrt{3x}+5\sqrt{27x}-16=\sqrt{432x}\) (*)
ĐKXĐ: \(x\ge0\)
(*) \(\Leftrightarrow\sqrt{3x}+5\sqrt{27x}-\sqrt{432x}=16\)
\(\Leftrightarrow\sqrt{3x}\left(1+5\sqrt{9}-\sqrt{144}\right)=16\)
\(\Leftrightarrow\sqrt{3x}\left(5\sqrt{9}+1-12\right)=16\)
\(\Leftrightarrow4\sqrt{3x}=16\)
\(\Leftrightarrow\sqrt{3x}=4\)
\(\Leftrightarrow3x=16\)
\(\Leftrightarrow x=\dfrac{16}{3}\) ( thỏa mãn đk )
Vậy \(s=\left\{\dfrac{16}{3}\right\}\)
