đặt \(\sqrt{4x+1}=a\) đk a>=0 =>\(a^2=4x+1\Leftrightarrow3a^2=12x+3\)
\(\sqrt{3x-2}=b\) đk b>=0 =>\(b^2=3x-2\Leftrightarrow4b^2=12x-8\)
=>\(\int^{a+b=5}_{3a^2-4b^2=11}\)\(\Leftrightarrow\int^{a=5-b}_{3a^2-4b^2=11}\)thay vào giải ra đc \(\int^{b=2}_{b=-32}\)mà b>=0=>b=2=>x=2