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b: 5x^2+5y^2+8xy-2x+2y+2=0
=>4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0
=>(x-1)^2+(y+1)^2+(2x+2y)^2=0
=>x=1 và y=-1
M=(1-1)^2015+(1-2)^2016+(-1+1)^2017=1
từ đề bài => \(x^2+2y+1+y^2+2z+1+z^2+2x+1=0\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)=> x=-1; y=-1 và z=-1
A=-1^2016+ -1^2016+ -1^2016=1+1+1=3
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0+0+0\)
\(\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
Mà \(\left(x+1\right)^2\ge0\)
\(\left(y+1\right)^2\ge0\)
\(\left(z+1\right)^2\ge0\)
\(\Rightarrow x+1=y+1=z+1=0\)
\(\Rightarrow x=y=z=-1\)
\(\Rightarrow P=1+1+1=3\)
cộng 3 vế lại cùng 1 lúc ta sẽ có (x+1)2 +(y+1)2+(z+1)2 = 0.
dấu bằng xảy ra khi cả 3 biểu thức bằng 0, suy ra x=y=z= -1
thế vào A thì A= -3
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Ta có: 5x2+5y2+8xy-2x+2y+2=0
=> 4x2+8xy+4y2+x2-2x+1+y2+2y+1=0
=> (2x+2y)2+(x-1)2+(y+1)2=0
=> {2x+2y=0 => x=-y
{x-1 = 0 => x=1
{y+1 =0 => y=-1
=> x=1, y=-1
Thay vào biểu thức M, ta có:
M=(1+-1)2015+(1-2)2016+(-1+1)2017=0+1+0=1 (đpcm)
x2 + 2y2 + z2 - 2xy - 2y - 4z + 5 = 0
<=> ( x2 - 2xy + y2 ) + ( y2 - 2y + 1 ) + ( z2 - 4z + 4 ) = 0
<=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2 = 0
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )2 + ( y - 1 )2 + ( z - 2 )2\(\ge\)0\(\forall\)x ; y ; z
Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )
Thay ( 1 ) vào A , ta được :
\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)
Vậy A = 2
Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)
Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)
Ta có : \(x^2+2y+1=0;y^2+2z+1=0;z^2+2x+1=0\)
\(\Rightarrow x^2+2y+1=y^2+2z+1=z^2+2x+1\)
\(\Rightarrow x^2+2y+1-y^2-2z-1-z^2-2x-1=0\)
\(\Rightarrow\left(x^2-2x+1\right)-\left(y^2-2y+1\right)-\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(x-1\right)^2-\left(y-1\right)^2-\left(z+1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-1=0\\z+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\\z=-1\end{cases}}\)
Thay \(x=1;y=1;z=-1\)vào A ta có :
\(A=1^{2015}+1^{2016}+\left(-1\right)^{2017}=1+1-1=1\)
Vậy A = 1
Từ \(\hept{\begin{cases}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{cases}}\)
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)
\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\left(1\right)\)
Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{cases}\left(2\right)}\)
Từ \(\left(1\right)\)và \(\left(2\right)\):
\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\\\left(z+1\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+1=0\\y+1=0\\z+1=0\end{cases}}\)
\(\Rightarrow x=y=z=-1\)
\(\Rightarrow A=\left(-1\right)^{2015}+\left(-1\right)^{2016}+\left(-1\right)^{2017}=-1+1-1=-1\)
Vậy \(A=-1\)