a) Ta có:
\(\sqrt{x-2}+\sqrt{3}=\sqrt{x}+1\)
\(\Leftrightarrow\left(\sqrt{x-2}+\sqrt{3}\right)^2=\left(\sqrt{x}+1\right)^2\)
\(\Leftrightarrow x-2+2\cdot\sqrt{x-2}\cdot\sqrt{3}+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow\)\(2\sqrt{3\left(x-2\right)}=2\sqrt{x}\)
\(\Leftrightarrow\sqrt{3x-6}=\sqrt{x}\)
\(\Leftrightarrow3x-6=x\)
<=> 2x-6=0
<=>x=3
Vậy.....