\(4x^2+3x+3=4x\sqrt{x+3}+2\sqrt{2x-1}\) ( ĐK : \(x\ge\frac{1}{2}\) )
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(\sqrt{2x-1}-1\right)^2=0\)
\(\Leftrightarrow x=1\)
Vậy...