\(Dk:x,y\ge\frac{-5}{4}\)
\(\left\{{}\begin{matrix}\left(2x-3\right)^2=4y+5\\\left(2y-3\right)^2=4x+5\end{matrix}\right.\Rightarrow\left(2y-3\right)^2-\left(2x-3\right)^2=4x-4y\Leftrightarrow\left(2y-2x\right)\left(2x+2y-6\right)=4\left(x-y\right)\Leftrightarrow4\left(y-x\right)\left(x+y-3\right)=4\left(x-y\right)\Leftrightarrow-4\left(x-y\right)\left(x+y-3\right)=4\left(x-y\right)\)
\(+,x=y\Rightarrow\left(2x-3\right)^2=4x+5\Leftrightarrow4x^2-12x+9=4x+5\Leftrightarrow4x^2-16x+4=0\Leftrightarrow x^2-4x+1=0\)
\(\Delta=16-4=12>0\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y=2+\sqrt{3}\left(tm\right)\\x=y=2-\sqrt{3}\left(tm\right)\end{matrix}\right.\)
\(+,x\ne y\Rightarrow-4\left(x+y-3\right)=4\Leftrightarrow x+y-3=-1\Leftrightarrow x+y=2\)
\(\Leftrightarrow x=2-y\Rightarrow\left(1-2y\right)^2=4y+5\Leftrightarrow1-4y+4y^2=4y+5\Leftrightarrow4y^2-8y-4=0\Leftrightarrow y^2-2y-1=0;\Delta=\left(-2\right)^2-\left(-1\right).1.4=4-\left(-4\right)=8>0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1-\sqrt{2};x=1+\sqrt{2}\left(tm\right)\\x=1-\sqrt{2};y=1+\sqrt{2}\left(tm\right)\end{matrix}\right.\)
