Điều kiện: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\y\ge\dfrac{3}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x\sqrt{y}+y\sqrt{x}=3\sqrt{4y-3}\left(1\right)\\2y\sqrt{x}+x\sqrt{y}=3\sqrt{4x-3}\left(2\right)\end{matrix}\right.\)
Lấy (1) - (2) ta được
\(x\sqrt{y}-y\sqrt{x}+3\left(\sqrt{4x-3}-\sqrt{4y-3}\right)=0\)
\(\Leftrightarrow\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\dfrac{12\left(x-y\right)}{\sqrt{4x-3}+\sqrt{4y-3}}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}+\dfrac{12\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{4x-3}+\sqrt{4y-3}}\right)=0\)
\(\Leftrightarrow x=y\)
\(\Rightarrow x\sqrt{x}=\sqrt{4x-3}\)
\(\Leftrightarrow x^3=4x-3\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-3\right)=0\)
Tới đây bí 
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\y\ge\dfrac{3}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2x\sqrt{y}+y\sqrt{x}=3\sqrt{4y-3}\left(1\right)\\2y\sqrt{x}+x\sqrt{y}=3\sqrt{4x-3}\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\) theo từng vế:
\(2x\sqrt{y}+y\sqrt{x}-2y\sqrt{x}-x\sqrt{y}=3\sqrt{4y-3}-3\sqrt{4x-3}\)
\(x\sqrt{y}-y\sqrt{x}-3\left(\sqrt{4y-3}-\sqrt{4x-3}\right)=0\)
\(\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\dfrac{12\left(x-y\right)}{\sqrt{4x-3}+\sqrt{4y-3}}=0\)
\(\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}+\dfrac{12\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{4x-3}-\sqrt{4y-3}}\right)=0\)
\(\Leftrightarrow x=y\)
\(\Rightarrow x\sqrt{x}=\sqrt{4x-3}\)
\(\Rightarrow x^3=4x-3\)
\(\left(x-1\right)\left(x^2+x+3\right)=0\)
Dễ thấy: \(x^2+x+3=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\forall x\)
\(\Rightarrow x=1\)
\(\Rightarrow y=1\)
