tìm x,y thuộc z để
2x^2+2xy+y^2-4x+2y+10=0
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tìm x;y thuộc z thỏa mãn x^2 + 2xy + 7(x+y) + 2y^2 + 10 = 0
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https://hoc24.vn/cau-hoi/tim-xy-thuoc-z-thoa-man-x2-2xy-7x-y-2y2-10-0.216670050813
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tìm x,y,z biết 2x^2+y^2-2xy+4x-2y=- 2(x,y,z thuộc Z+)
Tìm x,y,z biết: a) x^2+y^2-4x+4y+8=0 b) 5x^2-4xy+y^2=0 c) x^2+2y^2+z^2-2xy-2y-4z+5=0 d) 3x^2+3y^2+3xy-3x+3y+3=0 e) 2x^2+y^2+2z^2-2xy-2xz+2yz-2z-2z-2x+2=0
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
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
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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 \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
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d)3x2+3y2+3xy-3x+3y+3=0
⇔ 6x2+6y2+6xy-6x+6y+6=0
⇔ 3(x+y)2+3(x-1)2+3(y+1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
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Cho x y thuộc Z thỏa mãn
`x^2`+ `2xy` + `7x` + `7y` + `2y^2` + `10` = `0`
tìm gtnn và gtln của S= 2x+2y+2023
Tìm x, y thuộc Z để:
a) xy + x - y = 2
b) x - 2xy + y = 0
c) x. (x - 2) - (2 - x)y - 2. (x - 2) = 3
d) (2x - y). (4x2 + 2xy + y2) + (2x + y). (4x2 - 2xy + y2) - 16x. (x2 - y) = 32
e) x2 - 2xy + 2y2 - 2x + 6y +5 = 0
g) x2 + 2xy + 7x + 7y + 2y2 = 0
Tìm x, y thuộc Z để:
a) xy + x - y = 2
b) x - 2xy + y = 0
c) x. (x - 2) - (2 - x)y - 2. (x - 2) = 3
d) (2x - y). (4x2 + 2xy + y2) + (2x + y). (4x2 - 2xy + y2) - 16x. (x2 - y) = 32
e) x2 - 2xy + 2y2 - 2x + 6y +5 = 0
g) x2 + 2xy + 7x + 7y + 2y2 = 0
a) \(xy+x-y=2\)
\(\Leftrightarrow x\left(y+1\right)-\left(y+1\right)=1\)
\(\Leftrightarrow\left(x-1\right)\left(y+1\right)=1=1.1=\left(-1\right).\left(-1\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=y+1=1\\x-1=y+1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2;y=0\\x=0;y=-2\end{cases}}\)
b) \(x-2xy+y=0\)
\(\Leftrightarrow2x-4xy+2y=0\)
\(\Leftrightarrow2x\left(1-2y\right)-\left(1-2y\right)=-1\)
\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=-1\)
Tương tự nha
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c) \(x\left(x-2\right)-\left(2-x\right)y-2\left(x-2\right)=3\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)y-2\left(x-2\right)=3\)
\(\Leftrightarrow\left(x-2\right)\left(x+y-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=0\end{cases}}\)
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tim x y z biết
a,4x^2+9y^2+4x-24y+17=0
b,2x^2+2y^2+z^2+2xy-2xz-6y+9=0
c,x^2+2y+2xy+2x+6y+5=0
tim x y z biết
a,4x^2+9y^2+4x-24y+17=0
b,2x^2+2y^2+z^2+2xy-2xz-6y+9=0
c,x^2+2y+2xy+2x+6y+5=0
\(a,4x^2+9y^2+4x-24y+17=0\)
\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)
\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)
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\(\text{tìm x,y 2x^2 + y^2 - 2xy - 4x - 2y + 10 = 0}\)