Ta xét hệ \(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\left(1\right)\\x^2+y^2+x+y-4=0\left(2\right)\end{cases}}\)
Ta có: \(\left(1\right)\Leftrightarrow y^2-\left(x+1\right)y-2x^2+5x-2=0\)
\(\Leftrightarrow\left[y-\frac{x+1}{2}\right]^2-\left[\frac{\left(x+1\right)^2}{4}+2x^2-5x+2\right]=0\)
\(\Leftrightarrow\left[y-\frac{x+1}{2}\right]^2-\frac{9x^2-18x+9}{4}=0\)\(\Leftrightarrow\left[y-\frac{x+1}{2}\right]^2-\left(\frac{3x-3}{2}\right)^2=0\)
\(\Leftrightarrow\left(y-\frac{x+1}{2}-\frac{3x-3}{2}\right)\left(y-\frac{x+1}{2}+\frac{3x-3}{2}\right)=0\)\(\Leftrightarrow\left(y-2x+1\right)\left(y+x-2\right)=0\Leftrightarrow\orbr{\begin{cases}y-2x+1=0\\y+x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}y=2x-1\\y=2-x\end{cases}}\)
TH1: \(y=2x-1\), thay vào phương trình (2), ta được: \(x^2+\left(2x-1\right)^2+x+2x-1-4=0\)
\(\Leftrightarrow5x^2-x-4=0\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=1\\x=-\frac{4}{5}\Rightarrow y=\frac{-13}{5}\end{cases}}\)
TH2: \(y=2-x\), thay vào phương trình (2), ta được: \(x^2+\left(2-x\right)^2+x+2-x-4=0\)
\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow2\left(x-1\right)^2=0\Leftrightarrow x=1\Rightarrow y=1\)
Vậy hệ có 2 nghiệm \(\left(x;y\right)\in\left\{\left(1;1\right);\left(-\frac{4}{5};-\frac{13}{5}\right)\right\}\)
\(+,2x^2+xy-y^2-5x+y+2=0\)
\(\Leftrightarrow x^2+\frac{xy}{2}-\frac{y^2}{2}-\frac{5x}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow x^2+x\left(\frac{y}{2}-\frac{5}{2}\right)-\frac{y^2}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow x^2+2x.\frac{y-5}{4}+\left(\frac{y-5}{4}\right)^2-\left(\frac{y-5}{4}\right)^2-\frac{y^2}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}\right)^2-\frac{y^2-10y+25}{16}-\frac{y^2}{2}+\frac{y}{2}+1=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}\right)^2-\frac{9y^2-18y+9}{16}=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}\right)^2-\left(\frac{3y-3}{4}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{y-5}{4}-\frac{3y-3}{4}\right)\left(x+\frac{y-5}{4}+\frac{3y-3}{4}\right)=0\)
\(\Leftrightarrow\left(x+\frac{-y-1}{2}\right)\left(x+y+2\right)=0\)
\(\orbr{\begin{cases}x=\frac{y+1}{2}\\x=-y-2\end{cases}}\)
vậy ....
\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\left(1\right)\\x^2+y^2+x+y-4=0\left(2\right)\end{cases}}\)
PT (1) \(\Leftrightarrow2x^2+\left(5y-5\right)x-y^2+y+2=0\)
\(\Delta=\left(y-5\right)^2-8\left(-y^2+y+2\right)\)
\(=y^2-10y+25+8y^2-8y-16\)
\(=9y^2-18y+9\)
\(=\left(3y-3\right)^2\Rightarrow\sqrt{\Delta}=\left|3y-3\right|\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5-y+3y-3}{4}=\frac{2+2y}{4}=\frac{1+y}{2}\\x=\frac{5-y-3y+3}{4}=\frac{8-4y}{4}=2-y\end{cases}}\)
*) TH1: \(2x=1+y\)
=> y=-1+2x thay vào hệ phương trình (2) \(x^2+\left(2x-1\right)^2+x+2x-1-4=0\)
\(\Leftrightarrow x^2+4x^2-4x+1+3x-5=0\)
\(\Leftrightarrow5x^2-x-4=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-4}{5}\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}}\)và \(\hept{\begin{cases}x=\frac{-4}{5}\\y=\frac{-13}{5}\end{cases}}\)
*) TH2: \(x=2-y\Rightarrow y=2-x\)
=> PT(2) \(x^2+\left(2-x\right)^2+x+2-x-4=0\)
\(\Leftrightarrow x^2+x^2-4x+4-2=0\)
\(\Leftrightarrow2x^2+4x+2=0\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
<=> x=1
\(\Rightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)
Vậy hệ có nghiệm \(\left(x;y\right)=\left(1;1\right);\left(\frac{-4}{5};\frac{-13}{5}\right)\)
