ĐKXĐ: \(x\ge\frac23\)
Ta có: \(\sqrt{5x-1}+\sqrt{12x-8}=x^2+3\)
=>\(\sqrt{5x-1}-2+\sqrt{12x-8}-2=x^2+3-4\)
=>\(\frac{5x-1-4}{\sqrt{5x-1}+2}+\frac{12x-8-4}{\sqrt{12x-8}+4}=x^2-1=\left(x-1\right)\left(x+1\right)\)
=>\(\frac{5x-5}{\sqrt{5x-1}+2}+\frac{12x-12}{\sqrt{12x-8}+4}=\left(x-1\right)\left(x+1\right)\)
=>\(\left(x-1\right)\left(\frac{5}{\sqrt{5x-1}+2}+\frac{12}{\sqrt{12x-8}+4}-x-1\right)=0\)
=>x-1=0
=>x=1(nhận)
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