Ta có :
\(4x^2+3x+3=4x\sqrt{x+3}+2\sqrt{2x-1}\)
\(\Rightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1+2\sqrt{2x-1}+1\right)=0\)
\(\Rightarrow\left(2x-\sqrt{x+3}\right)^2+\left(\sqrt{2x-1}-1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-\sqrt{x+3}=0\\\sqrt{2x-1}-1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=\sqrt{x+3}\\2x-1=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=\sqrt{x+3}\\x=1\end{matrix}\right.\Rightarrow x=1\left(tm\right)\)
Vậy nghiệm của phương trình là x=1