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\(x^2+y^2-2xy+x-y+1=\left(x-y\right)^2+\left(x-y\right)+1\)
Đặt x-y=t
ta có: \(t^2+t+1=\left(t+\frac{1}{2}\right)^2+\frac{3}{4}>0,\forall t\)
=> \(x^2+y^2-2xy+x-y+1>0,\forall x,y\)
\(x^2+2xy+2y^2+y+\frac{1}{2}\)
\(=x^2+2xy+y^2+y^2+y+\frac{1}{2}\)
\(=\left(x+y\right)^2+y^2+y+\frac{1}{2}\)
Có : \(\left(x+y\right)^2\ge0\)
\(y^2\ge y\ge0\Rightarrow y^2+y\ge0\)
\(\frac{1}{2}>0\)
\(\Rightarrow x^2+2xy+2y^2+y+\frac{1}{2}>0\) với mọi x y
Ta có
\(x^2+2xy+2y^2+y+\frac{1}{2}\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}\)
\(=\left(x+y\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{2}\)
Mà \(\begin{cases}\left(x^2+2xy+y^2\right)\ge0\\\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)\ge0\\\frac{1}{4}>0\end{cases}\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(y^2+2.y.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}>0\)
\(x-2xy+y=0\)
\(\Rightarrow2x-4xy+2y=0\)
\(\Rightarrow2x-4xy+2y-1=-1\)
\(\Rightarrow\left(2x-4xy\right)-\left(1-2y\right)=-1\)
\(\Rightarrow2x.\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Rightarrow\left(2x-1\right).\left(1-2y\right)=-1.\)
Vì x, y là các số nguyên.
\(\Rightarrow\left(1-2y\right).\left(2x-1\right)\) là số nguyên.
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-2y=1\\2x-1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}1-2y=-1\\2x-1=1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2y=0\\2x=0\end{matrix}\right.\\\left\{{}\begin{matrix}2y=2\\2x=2\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\end{matrix}\right.\)
Vậy cặp số \(\left(x;y\right)\) thỏa mãn là: \(\left(0;0\right),\left(1;1\right).\)
Mình nghĩ thêm đề là tìm x, y nguyên.
Chúc bạn học tốt!
Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 M=x3+x2y−2x2−xy−y2+3y+x−1M=x3+x2y−2x2−xy−y2+3y+x−1
M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)M=x3+x2y−2x2−xy−y2+(2y+y)+x−(−2+1)
M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1M=(x3+x2y−2x2)−(xy+y2−2y)+(x+y−2)+1
M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1M=(x2.x+x2.y−2x2)−(x.y+y.y−2y)+(x+y−2)+1
M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1M=x2.(x+y−2)−y.(x+y−2)+(x+y−2)+1
M=x2.0+y.0+0+1M=x2.0+y.0+0+1
M=1M=1
N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−2
N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)N=x3+x2y−2x2−xy2+x2y+2xy+2y+2x−(−4+2)
N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2N=(x3+x2y−2x2)−(x2y+xy2−2xy)+(2x+2y−4)+2
N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2N=(x2x+x2y−2x2)−(xyx+xyy−2xy)+(2x+2y−4)+2
N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2N=x2(x+y−2)−xy(x+y−2)+2(x+y−2)+2
N=x2.0−xy.0+2.0+2N=x2.0−xy.0+2.0+2
N=2N=2
P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3P=x4+2x3y−2x3+x2y2−2x2y−x(x+y)+2x+3
P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x4+x3y−2x3)+(x3y+x2y2−2x2y)−(x2+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3P=(x3x+x3y−2x3)+(x2y.x+x2yy−2x2y)−(xx+xy−2x)+3
P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3P=x3(x+y−2)+x2y(x+y−2)−x(x+y−2)+3
P=x3.0+x2y.0−x.0+3P=x3.0+x2y.0−x.0+3
P=3
