Bài 2
a) \(\sqrt{x-1}+\sqrt{4x-4}=9\) (1)
ĐKXĐ: \(x\ge1\)
(1) \(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}=9\)
\(\Leftrightarrow3\sqrt{x-1}=9\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{9}{3}\)
\(\Leftrightarrow\sqrt{x-1}=3\)
\(\Leftrightarrow x-1=9\)
\(\Leftrightarrow x=10\) (nhận)
Vậy \(S=\left\{10\right\}\)
b) \(3\sqrt{2x-3}-6=9\) (2)
ĐKXĐ: \(2x-3\ge0\Leftrightarrow2x\ge3\Leftrightarrow x\ge\dfrac{3}{2}\)
(2) \(\Leftrightarrow3\sqrt{2x-3}=9+6\)
\(\Leftrightarrow3\sqrt{2x-3}=15\)
\(\Leftrightarrow\sqrt{2x-3}=\dfrac{15}{3}\)
\(\Leftrightarrow\sqrt{2x-3}=5\)
\(\Leftrightarrow2x-3=25\)
\(\Leftrightarrow2x=25+3\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=\dfrac{28}{2}\)
\(\Leftrightarrow x=14\) (nhận)
Vậy \(S=\left\{14\right\}\)