ĐKXĐ: \(2\le x\le\frac{10}{3}\)
Ta có: \(\sqrt{x-2}+\sqrt{10-3x}=5-x\)
=>\(\sqrt{x-2}-1+\sqrt{10-3x}-1=5-x-2\)
=>\(\frac{x-2-1}{\sqrt{x-2}+1}+\frac{10-3x-1}{\sqrt{10-3x}+1}=3-x\)
=>\(\frac{x-3}{\sqrt{x-2}+1}+\frac{9-3x}{\sqrt{10-3x}+1}=3-x\)
=>\(\left(x-3\right)\left(\frac{1}{\sqrt{x-2}+1}-\frac{3}{\sqrt{10-3x}+1}+1\right)=0\)
=>x-3=0
=>x=3(nhận)





