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\(x^2+4y^2+z^2-2x-6z+8y+15\)
\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)
\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1>0\forall x;y\)
\(x^2+5y^2+2x-4xy-10y+14\)
\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+y^2-6y+9+4\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4>0\forall x;y\)
Chúc bạn học tốt.
a.Ta có:\(2x^2-4xy+4y^2+2x+1=0\)
\(\Rightarrow\left[x^2-2x\left(2y\right)+\left(2y\right)^2\right]+\left(x^2+2x+1\right)=0\)
\(\Rightarrow\left(x-2y\right)^2+\left(x+1\right)^2=0\)
Dấu "=" xảy ra khi và chỉ khi x-2y=0 và x+1=0
Suy ra x=-1;y=-1/2
b.Ta có:\(x^2-6x+y^2-6y+21=3\)
\(\Rightarrow\left(x^2-6x+9\right)+\left(y^2-6y+9\right)+3-3=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y-3\right)^2=0\)
Dấu "=" xảy ra khi và chỉ khi x-3=y-3=0
Suy ra x=y=3
c.Ta có:\(2x^2-8x+y^2-2xy+16=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-4\right)^2=0\)
Dấu "=" xảy ra khi và chỉ khi:x-y=x-4=0
Suy ra x=y=4
a) 2x2 - 4xy + 4y2 + 2x + 1 = 0
<=> x2 - 4xy + 4y2 + x2 + 2x + 1 = 0
<=> ( x - 2y )2 + ( x + 1 )2 = 0
<=> \(\hept{\begin{cases}x-2y=0\\x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-\frac{1}{2}\end{cases}}\)
b) x2 - 6x + y2 - 6y + 21 = 3
<=> x2 - 6x + y2 - 6y + 21 - 3 = 0
<=> x2 - 6x + y2 - 6y + 18 = 0
<=> x2 - 6x + 9 + y2 - 6y + 9 = 0
<=> ( x - 3 )2 + ( y - 3 )2 = 0
<=> \(\hept{\begin{cases}x-3=0\\y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}\)
c) 2x2 - 8x + y2 - 2xy + 16 = 0
<=> x2 - 2xy + y2 + x2 - 8x + 16 = 0
<=> ( x - y )2 + ( x - 4 )2 = 0
<=> \(\hept{\begin{cases}x-y=0\\x-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=4\end{cases}}\)
1. Theo mình là sai đề, không biết có phải vậy không
2. (x^2 - 2.x.5 + 25) + (9y^2 - 2.3.2 +4) =0
(x-5)^2 + (3y-2)^2 = 0
TH1: (x-5)^2 = 0
x-5=0
x=5
TH2: (3y-2)^2 =0
3y -2=0
y=2/3
1. x2+y2-2x+4y+3=0
<=>(x2-2x+1)+(y2+4y+2)=0
<=>(x-1)2+(y+2)2=0
Mà \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
\(x^2+5y^2-4xy+2x-10y+14\)
\(=\left(x^2+4y^2-4xy+2x-4y+1\right)+\left(y^2-6y+9\right)+4\)
\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4\)
Vì \(\hept{\begin{cases}\left(x-2y+1\right)^2\ge0;\forall x,y\\\left(y-3\right)^2\ge0;\forall x,y\end{cases}}\)
\(\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2\ge0;\forall x,y\)
\(\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2+4\ge4>0;\forall x,y\)
Vậy ...
\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)
\(=3\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(3+2x-y\right)\left(2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)\)
\(=9x^2-\left(y-2\right)^2\)
\(=\left(3x+y-2\right)\left(3x-y+2\right)\)
A = ( 6x - 3y ) + (4x2 - 4xy + y2 )
A = 3.( 2x - y) + [ ( 2x )2 - 2.2.x.y + y2 ]
A = 3.( 2x - y ) + ( 2x - y )2
A = ( 2x - y ).(3 + 2x - y )
B = 9x2 - ( y2 - 4y + 4 )
B = ( 3x )2 - ( y - 2 )2
B = ( 3x - y + 2 ).( 3x + y - 2 )
C = - 25x2 + y2 - 6y + 9
C = ( y2 - 2.3.y + 32 ) - ( 5x )2
C = ( y - 3 )2 - ( 5x )2
C = (y - 3 - 5x ).( y - 3 +5x )
D = x2 - 4x - y2 -- 8y - 12
D = ( x2 - 4x + 4 ) - 4 - y2 - 8y -12
D = ( x - 2.2x + 22 ) - ( y2 + 2.4.y + 42 )
D = ( x - 2 )2 - ( y + 4 )2
D = ( x - 2 + y + 4 ).( x - 2 - y - 4 )
D = ( x + y + 2 ).( x - y - 6 )
Ta có : \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)
\(\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\Leftrightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
Do \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\\\left(y-3\right)^2\ge0\\\left(z-5\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=y+z\\y=3\\z=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}}\)
Khi đó \(P=\left(4-4\right)^{2018}+\left(3-4\right)^{2018}+\left(5-4\right)^{2018}\)
\(=0+\left(-1\right)^{2018}+1^{2018}\)
\(=2\)
b) 10x - x2 - 9y2 + 6y - 100
= - (x2 - 10x + 25) - (9y2 - 6y + 1) - 74
= - (x - 5)2 - (3x - 1)2 - 74 \(\le-74< 0\)