\(\hept{\begin{cases}2\sqrt{2xy-y}+2x+y=10\left(1\right)\\\sqrt{3y+4}-\sqrt{2y+1}+2\sqrt{2x-1}=3\left(2\right)\end{cases}}\)
\(ĐK:x\ge\frac{1}{2};y\ge0\)
\(\left(1\right)\Leftrightarrow\left(\sqrt{2x-1}+\sqrt{y}\right)^2=9\Leftrightarrow\sqrt{2x-1}+\sqrt{y}=3\)
\(\Leftrightarrow\sqrt{2x-1}=3-\sqrt{y}\)(*)
Thay \(\sqrt{2x-1}=3-\sqrt{y}\)vào (2), ta được: \(\sqrt{3y+4}-\sqrt{2y+1}-2\left(\sqrt{y}-2\right)-1=0\)
\(\Leftrightarrow\left(\sqrt{3y+4}-4\right)-\left(\sqrt{2y+1}-3\right)-2\left(\sqrt{y}-2\right)=0\)
\(\Leftrightarrow\frac{3\left(y-4\right)}{\sqrt{3y+4}+4}-\frac{2\left(y-4\right)}{\sqrt{2y+1}+3}-\frac{2\left(y-4\right)}{\sqrt{y}+2}=0\)
\(\Leftrightarrow\left(y-4\right)\left(\frac{3}{\sqrt{3y+4}+4}-\frac{2}{\sqrt{2y+1}+3}-\frac{2}{\sqrt{y}+2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y=4\Rightarrow x=1\\\frac{3}{\sqrt{3y+4}+4}=\frac{2}{\sqrt{2y+1}+3}+\frac{2}{\sqrt{y}+2}\left(3\right)\end{cases}}\)
Với \(y\ge0\)thì \(\frac{3}{\sqrt{3y+4}+4}\le\frac{1}{2}\)
Từ (*) suy ra \(y\le9\Rightarrow\frac{2}{\sqrt{2y+1}+3}+\frac{2}{\sqrt{y}+2}>\frac{1}{2}\)
Suy ra (3) vô nghiệm
Vậy hệ có cặp nghiệm duy nhất \(\left(x,y\right)=\left(1,4\right)\)
