Đặt \(\sqrt{4x+1}=a\)
\(\sqrt{3x-2}=b\)
=>\(9\left(a-b\right)=a^2-b^2\)
=>\(9=a+b\)
Ta có a+b=9
=> \(\sqrt{4x+1}+\sqrt{3x-2}=9\)9
=>\(\sqrt{4x+1}-5+\sqrt{3x-2}-4=0\)
=>\(\frac{4x+1-25}{\sqrt{4x+1}+5}+\frac{3x-2-16}{\sqrt{3x-2}+4}=0\)
=>\(\frac{4x-24}{\sqrt{4x+1}+5}+\frac{3x-18}{\sqrt{3x-2}+4}=0\)
=>\(\left(x-6\right)\left(\frac{4}{\sqrt{4x+1}+5}+\frac{3}{\sqrt{3x-2}+4}\right)=0\)
=>x-6=0
=>x=6