1.\(\left\{{}\begin{matrix}x^2+2xy-2x-y=0\\x^4-4\left(x+y-1\right)x^2+y^2+2xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y\right)\left(x-1\right)=0\\x^4-4\left(x+y-1\right)x^2+y^2+2xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1^4-4\left(1+y-1\right)1^2+y^2+2.1.y=0\end{matrix}\right.\)(1)
hoặc \(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^4-4\left(x-2x-1\right)x^2+\left(-2x\right)^2+2x.\left(-2x\right)=0\end{matrix}\right.\)(2)
(1)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\1-4y+y^2+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y^2-2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
(2)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^4-4\left(-x-1\right)x^2+4x^2-4x^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^2\left(x^2+4x+4\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x^2\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}y=4\\x=-2\end{matrix}\right.\)
Vậy nghiệm của hệ pt là (1;1),(0;0),(-2;4)
2. \(x^4-x^3+1-y^2=0\)
\(\Leftrightarrow x^3\left(x-1\right)+\left(1-y\right)\left(1+y\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3\left(x-1\right)=0\\\left(1-y\right)\left(1+y\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)(tm)hoặc\(\left\{{}\begin{matrix}x=1\\y=\pm1\end{matrix}\right.\)(tm)
Vậy nghiệm nguyên cuar pt là (0;1),(0;-1),(1;1),(1;-1)
Câu 1:
\(\left\{\begin{matrix} x^2+2xy-2x-y=0(1)\\ x^4-4(x+y-1)x^2+y^2+2xy=0(2)\end{matrix}\right.\)
Bình phương (1)
\((x^2+2xy-2x-y)^2=0\)
\(\Leftrightarrow (x^2+2xy)^2+(2x+y)^2-2(x^2+2xy)(2x+y)=0(3)\)
Lấy \((3)-(2)\) thu được:
\(4x^3y+4x^2y^2-6x^2y-4xy^2+2xy=0\)
\(\Leftrightarrow 2xy[2x^2+2xy-3x-2y+1]=0\)
\(\Leftrightarrow 2xy[2x(x-1)+2y(x-1)-(x-1)]=0\)
\(\Leftrightarrow 2xy(2x+2y-1)(x-1)=0\)
Do đó xét các TH sau:
TH1: \(x=0\) thay vào (1) suy ra \(y=0\)
TH2: \(y=0\Rightarrow x^2-2x=0\Leftrightarrow x=0;2\)
TH3: \(x=1\). Thay vào (1) suy ra \(y=1\). Thử lại thấy đúng.
TH4: \(2x+2y-1=0\)
\((1)\Rightarrow (x+y-1)^2=y^2-y+1\)
\(\Leftrightarrow y^2-y+1=(\frac{1}{2}-1)^2=\frac{1}{4}\)
\(\Leftrightarrow y^2-y+\frac{3}{4}=0\)
\(\Leftrightarrow (y-\frac{1}{2})^2+\frac{1}{2}=0\) (vô lý)
Vậy \((x,y)=(0,0); (2,0); (1,1)\)
