\(A=\left(4x^2-4xy+y^2\right)+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{21}{4}\\ A=\left(2x-y\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\\ A_{min}=-\dfrac{21}{4}\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-3\end{matrix}\right.\)

\(B=\left[\left(x-1\right)\left(x+2\right)\right]\left[x\left(x+1\right)\right]=\left(x^2+x-2\right)\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)\\ B=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-1=\left(x^2+x-1\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow x^2+x-1=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\\ \Leftrightarrow\left(x+\dfrac{1-\sqrt{5}}{2}\right)\left(x+\dfrac{1+\sqrt{5}}{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

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

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=5x2+2y2−4xy−8x−4y+19=(2x2−4xy+2y2)+4(x−y)+(3x2−12x)+19=2(x−y)2+4(x−y)+3(x2−4x+4)+7=2[(x−y)2+2(x−y)+1]+3(x−2)2+5=2(x−y+1)2+3(x−2)2+5≥0Dấu "=" xảy ra khi{x−y+1=0x−2=0↔{x=2y=x+1=3VậyMinA=5↔{x=2y=3

mik viết 5x2 là 5x mũ 2 nha

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

B=5x²+4xy-2(x-2y)+2y²+3

=5x²+4xy-2x+4y+2y²+3

=(4x²+4xy+y²)+(x²-2x+1)+(y²+4y+4)-2

=(2x+y)²+(x-1)²+(y+2)²-2  \(\ge\) -2

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x+y=0\\x-1=0\\y+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)

thanks b

b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)

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

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

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

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

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

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

\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)