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\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
\(B=\left(x^2+y^2+4+2xy-4x-4y\right)+\left(x^2+z^2+1+2xz-2x-2z\right)+\left(y^2-4y+4\right)+4\)
\(B=\left(x+y-2\right)^2+\left(x+z-1\right)^2+\left(y-2\right)^2+4\ge4\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x+y-2=0\\x+z-1=0\\y-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\\z=1\end{matrix}\right.\)
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$
$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$
$\Leftrightarrow x=3; y=-2$
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$B=9x^2+y^2+2z^2-18x+4z-6y+30$
$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$
$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$
$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
Vậy $B_{\min}=10$. Giá trị này đạt tại $(x-1)^2=(y-3)^2=(z+1)^2$
$\Leftrightarrow x=1; y=3; z=-1$
$C=x^2+y^2+z^2-xy-yz-xz+3$
$2C=2x^2+2y^2+2z^2-2xy-2yz-2xz+6$
$=(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)+6$
$=(x-y)^2+(y-z)^2+(z-x)^2+6\geq 6$
$\Rightarrow C\geq 3$
Vậy $C_{\min}=3$. Giá trị này đạt tại $x-y=y-z=z-x=0$
$\Leftrihgtarrow x=y=z$
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$D=5x^2+2y^2+4xy-2x+4y+2021$
$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$
$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$
$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$
$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$
Vậy $D_{\min}=2016$ khi $x+y+1=x-1=0$
$\Leftrightarrow x=1; y=-2$
\(Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5=\left[\left(x^2+2xy+y^2\right)-2z\left(x+y\right)+z^2\right]+\left(y^2-2y+1\right)+\left(z^2+4z+4\right)=\left(x+y-z\right)^2+\left(y-1\right)^2+\left(z+2\right)^2\ge0\)
\(minQ=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-3\\y=1\\z=-2\end{matrix}\right.\)
`Q=x^2+2y^2+2z^2+2xy-2yz-2xz-2y+4z+5`
`Q=(x^2+y^2-z^2+2xy-2yz-2xz)+(y^2-2y+1)+(z^2+4z+4)`
`Q=(x+y-z)^2+(y-1)^2+(z+2)^2`
Ta thấy :
`(x+y-z)^2>=0`
`(y-1)^2>=0`
`(z+2)^2>=0`
`=>(x+y-z)^2+(y-1)^2+(z+2)^2>=0`
Dấu = xảy ra
`<=>` $\begin{cases}x+y-z=0\\y-1=0\\z+2=0\end{cases}$
`<=>` $\begin{cases}x=-3\\y=1\\z=-2\end{cases}$
a) \(\Leftrightarrow4x^2+2y^2+4xy-20x-8y+26=0\)
\(\Leftrightarrow4x^2+4x\left(y-5\right)+\left(y-5\right)^2-\left(y-5\right)^2+2y^2-8y+26=0\)
\(\Leftrightarrow\left(2x+y-5\right)^2+y^2+2y+1=0\)
\(\Leftrightarrow\left(2x+y-5\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-5=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\) ( TM )
b) \(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)+\left(z^2-2z+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^2+\left(z-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\\z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\\z=1\end{matrix}\right.\) ( TM )
c) \(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+2x+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+1\right)^2+\left(z-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x+1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-1\\z=2\end{matrix}\right.\) ( TM )
\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)
