\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}y^2-\left(x+1\right)y-2x^2+5x-2=0\\x^2+y^2+x+y-4=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(y+x-2\right)\left(y-2x+1\right)=0\\x^2+y^2+x+y-4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+x-2=0\\x^2+y^2+x+y-4=0\end{cases}}\)hoặc \(\hept{\begin{cases}y-2x+1=0\\x^2+y^2+x+y-4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)hoặc \(\hept{\begin{cases}x=\frac{-4}{5}\\y=\frac{-13}{5}\end{cases}}\)và \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)

Vậy hpt có 2 nghiệm (x;y)=\(\left(1;1\right);\left(\frac{-4}{5};\frac{-13}{5}\right)\)

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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 \(\left(x;y\right)=\left(1;1\right);\left(\frac{-4}{5};\frac{-13}{5}\right)\)

\(\hept{\begin{cases}2x^2+xy-y^2-5x+y+2=0\\x^2+y^2+x+y-4=0\end{cases}}\)

\(\Leftrightarrow2x^2+xy-y^2-5x+y+2=x^2+y^2+x+y-4\)

\(\Leftrightarrow x^2+xy-y^2-5x+y+2=y^2+x+y-4\)

\(\Leftrightarrow x^2+xy-y^2-5x+y=y^2+x+y-4-2\)

\(\Leftrightarrow x^2+xy-y^2-5x+y=y^2+x+y-6\)

\(\Leftrightarrow x^2+xy-y^2+y=y^2+x+y-6+5x\)

\(\Leftrightarrow x^2+xy-y^2+y=y^2+6x+y-6\)

\(\Leftrightarrow x^2+xy-y^2=y^2+6x-6\)

\(\Leftrightarrow x^2+xy=y^2+6x-6+y^2\)

\(\Leftrightarrow x^2+xy=2y^2+6x-6\)

\(\Leftrightarrow x\left(x+y\right)=2\left(y^2+3x-3\right)\)

pt (1) <=>5x-2x^2-xy+y^2-y-2=0 

giai phuong trinh (1) theo an y ta co: 
y² - (x+1)y - (2x² - 5x+2)=0 
<=>Δ=(x+1)²+4(2x² - 5x+2)=x²+2x+1+8x²-20y+8=9x²-18x+9 
=9(x-1)² 
Δ>=0 => phuong trinh co nghiem 
<=>y=(x+1+3(x-1))/2 hoac y=(x+1-3(x-1))/2 
<=>y=2x-1 hoac y=2-x 
* thay y=2x-1 vao pt 2 ta duoc: 
x²+(2x-1)²+x+(2x-1)=4 
<=>5x²-x-4=0 
giai phuong trinh tren ta tim duoc x=1 va y=1 hoac x=-4/5 va y=-13/5 
*the y=2-x vao pt 2 ta duoc 
x²+(2-x)²+x+(2-x)=4 
<=>2x²-4x+2=0 
<=>x=1 =>y=1 
vay phuong trinh co 2 nghiem (1;1);(-4/5;-13/5)

\(pt\left(1\right)\Leftrightarrow\left(x+y-2\right)\left(2x-y-1\right)=0\)

Chia 2 trường hợp vào dùng pp thế, thế xuống pt dưới.

\(\left\{{}\begin{matrix}2x^2+xy-y^2-5x+y+2=0\left(1\right)\\x^3+y^2+x+y-4=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2x^2+2xy-4x\right)+\left(-xy-y^2+2y\right)+\left(-x-y+2\right)=0\)

\(\Leftrightarrow\left(x+y-2\right)\left(2x-y-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2-x\left(3\right)\\y=2x-1\left(4\right)\end{matrix}\right.\)

Thế (3) vào (2) ta được:

\(x^3+\left(2-x\right)^2+x+\left(2-x\right)-4=0\)

\(\Leftrightarrow x^3+x^2-4x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}-1\\x=-\sqrt{3}-1\end{matrix}\right.\)

Tương tự cho trường hợp còn lại.

\(pt\left(1\right)\Leftrightarrow\left(2x-y-1\right)\left(x+y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-y-1=0\\x+y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=2x-1\\y=2-x\end{matrix}\right.\)

*)Xét \(y=2x-1\) thì

\(pt\left(2\right)\Leftrightarrow x^3+\left(2x-1\right)^2+x+\left(2x-1\right)-4=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=1\\x=-1\end{matrix}\right.\)

*)Xét \(y=2-x\) thì:

\(pt\left(2\right)\Leftrightarrow x^3+\left(2-x\right)^2+x+\left(2-x\right)-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x^2+2x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-2\pm\sqrt{12}}{2}\end{matrix}\right.\)

\(y^2-\left(x+1\right)y-2x^2+5x+2=0\Leftrightarrow\left(y+x-2+1\right)=0.\)Do đó hệ đã cho tương đương với:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\x^2+y^2+x+y-4=0\end{matrix}\right.\\\left\{{}\begin{matrix}y-2x+1=0\\x^2+y^2+x+y-4=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{4}{5}\\y=-\dfrac{13}{5}.\end{matrix}\right.\end{matrix}\right.\)