Điều kiện: x \(\ge\frac{5}{3}\)
PT <=> \(\sqrt{8x+1}-\sqrt{7x+4}=\sqrt{2x-2}-\sqrt{3x-5}\)
<=> \(\frac{\left(8x+1\right)-\left(7x+4\right)}{\sqrt{8x+1}+\sqrt{7x+4}}=\frac{\left(2x-2\right)-\left(3x-5\right)}{\sqrt{2x-2}+\sqrt{3x-5}}\) <=> \(\frac{x-3}{\sqrt{8x+1}+\sqrt{7x+4}}=\frac{-\left(x-3\right)}{\sqrt{2x-2}+\sqrt{3x-5}}\)
<=> \(\frac{x-3}{\sqrt{8x+1}+\sqrt{7x+4}}+\frac{x-3}{\sqrt{2x-2}+\sqrt{3x-5}}=0\)
<=> \(\left(x-3\right)\left(\frac{1}{\sqrt{8x+1}+\sqrt{7x+4}}+\frac{1}{\sqrt{2x-2}+\sqrt{3x-5}}\right)=0\)
<=> x - 3 = 0 (Do \(\frac{1}{\sqrt{8x+1}+\sqrt{7x+4}}+\frac{1}{\sqrt{2x-2}+\sqrt{3x-5}}>0\) với mọi x > =5/3)
<=> x = 3 ( T/m)
Vậy..............
