Ta có:

\(3-S=\left(x^2+4y^2+9z^2\right)-\left(2x+4y+6z\right)\)

\(\Leftrightarrow3-S=\left(x^2-2x+1\right)+\left(4y^2-4y+1\right)+\left(9z^2-6z+1\right)-3\)

\(\Leftrightarrow6-S=\left(x-1\right)^2+\left(2y-1\right)^2+\left(3z-1\right)^2\ge0\)

\(\Leftrightarrow S\le6\)

\(S_{max}=6\) khi \(\left\{{}\begin{matrix}x-1=0\\2y-1=0\\3z-1=0\end{matrix}\right.\) \(\Leftrightarrow\left(x;y;z\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)

$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$

$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
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 $x+y+1=x-1=0$

$\Leftrightarrow x=1; y=-2$

$E=x^2-2x+4y^2+4y+2014$

$=(x^2-2x+1)+(4y^2+4y+1)+2012$

$=(x-1)^2+(2y+1)^2+2012$

$\geq 2012$

Vậy $E_{\min}=2012$. Giá trị này đạt tại $x-1=2y+1=0$

$\Leftrightarrow x=1; y=\frac{-1}{2}$

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$F=5x^2+5y^2+8xy+2y-2x+30$

$=4(x^2+2xy+y^2)+x^2+y^2+2y-2x+30$

$=4(x+y)^2+(x^2-2x+1)+(y^2+2y+1)+28$

$=4(x+y)^2+(x-1)^2+(y+1)^2+28\geq 28$

Vậy $F_{\min}=28$. Giá trị này đạt tại $x+y=x-1=y+1=0$

$\Leftrightarrow x=1; y=-1$

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

`a)x^2-2x+2+4y^2+4y`

`=x^2-2x+1+4y^2+4y+1`

`=(x-1)^2+(2y+1)^2`

`b)4x^2+y^2+12x+4y+13`

`=4x^2+12x+9+y^2+4y+4`

`=(2x+3)^2+(y+2)^2`

`c)x^2+17+4y^2+8x+4y`

`=x^2+8x+16+4y^2+4y+1`

`=(x+4)^2+(2y+1)^2`

`d)4x^2-12xy+y^2-4y+13`

`=4x^2-12x+9+y^2-4y+4`

`=(2x-3)^2+(y-2)^2`

a) \(x^2-2x+2+4y^2+4y=\left(x-1\right)^2+\left(2y+1\right)^2\)

b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)

c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)

d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)

a: \(x^2-2x+2+4y^2+4y\)

\(=x^2-2x+1+4y^2+4y+1\)

\(=\left(x-1\right)^2+\left(2y+1\right)^2\)

b: \(4x^2+12x+y^2+4y+13\)

\(=4x^2+12x+9+y^2+4y+4\)

\(=\left(2x+3\right)^2+\left(y+2\right)^2\)

c: \(x^2+8x+4y^2+4y+17\)

\(=x^2+8x+16+4y^2+4y+1\)

\(=\left(x+4\right)^2+\left(2y+1\right)^2\)

d: \(4x^2-12x+y^2-4y+13\)

\(=4x^2-12x+9+y^2-4y+4\)

\(=\left(2x-3\right)^2+\left(y-2\right)^2\)

\(P=x^2+4xy+4y^2-4xy-4y^2+2x+3\)

\(=x^2+2x+3\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

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