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x2 - 2xy + 2y2 - 2x + 6y + 13 = 0
<=> x2 - 2x(y + 1) + 2y2 + 6y + 13 = 0
<=> x2 - 2x(y + 1) + (y + 1)2 + y2 + 4y + 12 = 0
<=> (x - y - 1)2 + (y + 1)2 + (y + 2)2 + 8 = 0
Vô lí do VT > 0 vs mọi x; y
=> Ko tìm đc gtri của N
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\(a)xy+3x-2y=11\)
\(\Leftrightarrow xy+3x-2y-6=5\)
\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)
\(b)2x^2-2xy+x-y=12\)
\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)
\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)
\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)
\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)
Vì 2x+1 luôn lẻ
\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
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Lời giải:
Ta có:
\(x^2-2x+2y^2-2x-2y+5=0\)
\(\Leftrightarrow (x^2+y^2+1-2xy-2x+2y)+(y^2-4y+4)=0\)
\(\Leftrightarrow (x-y-1)^2+(y-2)^2=0(*)\)
Vì \((x-y-1)^2, (y-2)^2\geq 0, \forall x,y\in\mathbb{Z}\) nên $(*)$ xảy ra khi và chỉ khi:
\(\left\{\begin{matrix} (x-y-1)^2=0\\ (y-2)^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y-1=0\\ y=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=3\\ y=2\end{matrix}\right.\)
Do đó thay các giá trị cụ thể của $x,y$ vào biểu thức $P$ thì:
\(P=1\)
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a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
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a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)