ĐKXĐ: \(2\le x\le\frac{10}{3}\)
\(\sqrt{x-2}+\sqrt{10-3x}=5-x\)
\(\Rightarrow2\sqrt{x-2}+2\sqrt{10-3x}=10-2x\)
\(\Rightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(10-3x-2\sqrt{10-3x}+1\right)=0\)
\(\Rightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{10-3x}-1\right)^2=0\)
\(\Rightarrow\sqrt{x-2}-1=0,\sqrt{10-3x}-1=0\)
\(\Rightarrow x=3\)
Vậy \(x=3\)