\(x^2+2xy+y^2+3y-4=0\)
\(\Rightarrow\Delta'=y^2-\left(2y^2+3y-4\right)\ge0\)
\(\Leftrightarrow-4\le y\le1\)
\(\left(x+y\right)^2+\left(y-\frac{3}{2}\right)^2=4\)
mà 4=0^2+2^2
=>\(\orbr{\begin{cases}\hept{\begin{cases}x+y=0\\y-\frac{3}{2}=2\end{cases}}\\\hept{\begin{cases}x+y=2\\y-\frac{3}{2}=0\end{cases}}\end{cases}}\)
=> giải nốt
\(x^2+2y^2+2xy+3y-4=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2+3y+\frac{9}{4}\right)-\frac{25}{4}=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y+\frac{3}{2}\right)^2=\frac{25}{4}\)
\(\Leftrightarrow\left(2x+2y\right)^2+\left(2y+3\right)^2=25\)
Ta có 4 trường hợp:
TH1: \(\hept{\begin{cases}2x+2y=0\\2y+3=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)
TH2: \(\hept{\begin{cases}2x+2y=0\\2y+3=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)
TH3: \(\hept{\begin{cases}2x+2y=4\\2y+3=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=-3\end{cases}}\)
TH4: \(\hept{\begin{cases}2x+2y=4\\2y+3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)
TH5: \(\hept{\begin{cases}2x+2y=-4\\2y+3=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
TH6: \(\hept{\begin{cases}2x+2y=-4\\2y+3=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-2\\y=0\end{cases}}\)
