a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)

\(=x^2-2xy+y^2+4y^2+4y+1+50\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)

a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)

\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)

c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)

\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)

d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)

\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)

\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)

a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)

b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)

\(M=x^2-2xy+4y^2+12xy+22\)

\(M=\left(x^2-2xy+y^2\right)+\left(3y^2+12y+12\right)+10\)

\(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=-2\) 

( Chỗ \(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\) bạn phân tích từng cái đã nhá, mình làm tắt ) 

`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: \(4x^2-12x+y^2-4y+13\)

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

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

mk trả lời cho bạn rồi mà

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

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

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

\(A_{\left(x,y\right)}=x^2+4y^2+1-4xy+2x-4y\)

Đặt 2y=z

\(A_{\left(x,z\right)}=x^2+z^2+1-2xz+2x-2z\)

\(A_{\left(x,z\right)}=\left(x^2-xz\right)+\left(z^2-xz\right)+\left(x-z\right)+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left[x\left(x-z\right)+z\left(z-x\right)+\left(x-z\right)\right]+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left[\left(x-z\right)\left(x-z+1\right)\right]+\left(x-z+1\right)\)

\(A_{\left(x,z\right)}=\left(x-z+1\right)\left(x-z+1\right)=\left(x-z+1\right)^2\)

Vậy nghiệm đã thức là: \(x-z+1=0\Leftrightarrow x-2y+1=0\)

p/s: lớp 8 không dài dòng thế này%

Phân tích đa thức thành nhân tử

1: \(x^2-x-y^2-y\)

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

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

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

2: \(x^2-y^2+x-y\)

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

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

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

3: \(3x-3y+x^2-y^2\)

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

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

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

4: \(5x-5y+x^2-y^2\)

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

\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(5+x+y\right)\)

5: \(x^2-5x-y^2-5y\)

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

\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-5\right)\)

6: \(x^2-y^2+2x-2y\)

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

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

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

7: \(x^2-4y^2+x+2y\)

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

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

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

8: \(x^2-y^2-2x-2y\)

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

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

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

9: \(x^2-4y^2+2x+4y\)

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

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

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