ĐKXĐ: \(\begin{cases}7-x\ge0\\ x-5\ge0\end{cases}\Rightarrow5\le x\le7\)
Ta có: \(\sqrt{7-x}+\sqrt{x-5}=x^2-12x+38\)
=>\(\sqrt{7-x}-1+\sqrt{x-5}-1=x^2-12x+36\)
=>\(\frac{7-x-1}{\sqrt{7-x}+1}+\frac{x-5-1}{\sqrt{x-5}+1}=\left(x-6\right)^2\)
=>\(\left(x-6\right)\left(\frac{-1}{\sqrt{7-x}+1}+\frac{1}{\sqrt{x-5}+1}\right)-\left(x-6\right)^2=0\)
=>\(\left(x-6\right)\left(\frac{-1}{\sqrt{7-x}+1}+\frac{1}{\sqrt{x-5}+1}-x+6\right)=0\)
=>x-6=0
=>x=6(nhận)
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