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

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

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

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

\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

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

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

⇔(x−1)^2+(2y−3)^2+(z+2)^2=0

\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

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

\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương

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

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

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

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

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

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

\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

a,=x(x−2)b,=2b(x−3y)+a(x−3y)=(a+2b)(x−3y)c,=x(x2+2xy+y2−4)=x[(x+y)2−4]=x(x+y+2)(x+y−2)d,=4−(x+y)2=(2−x−y)(2+x+y)đ,=5(x−y)(x+y)+3(x+y)2=(x+y)(5x−5y+3x+3y)=(x+y)(8x−2y)=2(4x−y)(x+y)e,=3x(2xy−3)b,=x(4x2−4xy+y2−4)=x[(2x−y)2−4]=x(2x−y−2)(2x−y+2)f,=(x+y)2−z2=(x+y−z)(x+y+z)

\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)

a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

\(minA=2\Leftrightarrow x=3\)

b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)

\(minB=51\Leftrightarrow x=5\)

c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

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

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3