Bài 2:
a: Ta có: \(\sqrt{64x+64}-\sqrt{25x+25}+\sqrt{4x+4}=20\)
\(\Leftrightarrow8\sqrt{x+1}-5\sqrt{x+1}+2\sqrt{x+1}=20\)
\(\Leftrightarrow5\sqrt{x+1}=20\)
\(\Leftrightarrow x+1=16\)
hay x=15
b: Ta có: \(\sqrt{3x}+5\sqrt{27x}-16=\sqrt{432x}\)
\(\Leftrightarrow\sqrt{3x}+15\sqrt{3x}-12\sqrt{3x}=16\)
\(\Leftrightarrow4\sqrt{3x}=16\)
\(\Leftrightarrow3x=16\)
hay \(x=\dfrac{16}{3}\)
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