\(\sqrt{x-2}+\sqrt{x+1}=3\)
\(\Leftrightarrow\sqrt{x+1-3}+\sqrt{x+1}=3\)
Đặt \(t=x+1\left(t\ge3\right)\)
\(pt\Leftrightarrow\sqrt{t-3}+\sqrt{t}=3\)
\(\Leftrightarrow\sqrt{t-3}=3-\sqrt{t}\)
\(\Leftrightarrow t-3=t-6\sqrt{t}+9\)
\(\Leftrightarrow6\sqrt{t}=12\Leftrightarrow\sqrt{t}=2\Leftrightarrow t=4\)
=> x + 1 = 4 <=> x = 3 (t/m)
Vậy pt có nghiệm x = 3