ĐKXĐ: x>=2/3
Ta có: \(\sqrt{x}+\sqrt{3x-2}=x^2+1\)
=>\(\sqrt{x}-1+\sqrt{3x-2}-1=x^2+1-2\)
=>\(\frac{x-1}{\sqrt{x}+1}+\frac{3x-2-1}{\sqrt{3x-2}+1}=\left(x-1\right)\left(x+1\right)\)
=>\(\left(x-1\right)\left(\frac{1}{\sqrt{x}+1}+\frac{3}{\sqrt{3x-2}+1}-x-1\right)=0\)
=>x-1=0
=>x=1(nhận)

