\(đkx\ge0\\ 2x+5\sqrt{x}-3=0\\ =>2x-\sqrt{x}+6\sqrt{x}-3=0\\ =>\sqrt{x}\left(2\sqrt{x}-1\right)+3\left(2\sqrt{x}-1\right)=0\\ =>\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}+3=0\\2\sqrt{x}-1=0\end{matrix}\right.=>\left[{}\begin{matrix}\sqrt{x}=-3\left(voli\right)\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\\ =>x=\dfrac{1}{4}\left(thoaman\right)\)
\(2x+5\sqrt{x}=3\left(x\ge0\right)\)
\(\Leftrightarrow2x+5\sqrt{x}-3=0\)
\(\Leftrightarrow2x-\sqrt{x}+6\sqrt{x}-3=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)+3\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x}+3=0\\2\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(vô.lí\right)\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(tm\right)\\x=-\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)
