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

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

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

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

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

$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$

$\Leftrightarrow x=3; y=-2$

---------------------

$B=9x^2+y^2+2z^2-18x+4z-6y+30$

$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$

$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$

$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
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 $x-y=y-z=z-x=0$

$\Leftrihgtarrow x=y=z$

--------------------------------------

$D=5x^2+2y^2+4xy-2x+4y+2021$

$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$

$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$

$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$

$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$

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 $x-1=2y+1=0$

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

----------------------

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

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

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

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

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

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

a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x-6y-1\right)\)

b) \(=x\left(x^2-8x+16\right)\)

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

c) \(=2\left(x-y\right)^2-18\)

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

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

a: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: \(x^3-8x^2+16x\)

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

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

c: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

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

d: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)

\(=\left(x^2-6\right)\left(x^2+5\right)\)

f: Ta có: \(x^2-xy-2y^2\)

\(=x^2-2xy+xy-2y^2\)

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

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

g: Ta có: \(x^4-13x^2y^2+4y^4\)

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

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

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

a: Ta có: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: Ta có: \(16x-8x^2+x^3\)

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

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

c: Ta có: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

\(=2\cdot\left[\left(x-y\right)^2-9\right]\)

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

d: Ta có: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)

\(=\left(x^2-6\right)\left(x^2+5\right)\)

f: Ta có: \(x^2-xy-2y^2\)

\(=x^2-2xy+xy-2y^2\)

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

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

g: Ta có: \(x^4-13x^2y^2+4y^4\)

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

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

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

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

\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)

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

h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)

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

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

\(=\left(x^2-2x-3\right)\left(x^2-2x+1\right)\)

\(=\left(x-3\right)\left(x+1\right)\cdot\left(x-1\right)^2\)

a, xem lại đề 

\(b,x^2-4x+y^2-6y+1\\ =\left(x^2-4x+4\right)+\left(y^2-6y+9\right)-12\\ =\left(x-2\right)^2+\left(y-3\right)^2-12\ge-12\)

Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy ...

\(c,x^2-4xy+5y^2-2y+5\\ =\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+4\\ =\left(x-2y\right)^2+\left(y-1\right)^2+4\ge4\)

Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy ...

a, 

Dấu "=" xảy ra⇔{x=2y=3⇔{x=2y=3

Vậy ...

Dấu "=" xảy ra⇔{x=2y=1⇔{x=2y=1

Vậy ...

x² - 2xy + 2y² - 2x + 6y + 13 = 0 

<=> x² - 2x(y + 1) + 2y² + 6y + 13 = 0 

<=> x² - 2x(y + 1) + (y + 1)² + y² + 4y + 12 = 0 

<=> (x - y - 1)² + (y + 1)² + (y + 2)² + 8 = 0 

Vô lí do VT > 0 vs mọi x; y 

=> Ko tìm đc gtri của N

khongg lam maa ddoi co an thi an cai lon me may