a)  \(A=x^2+2xy+y^2-4x-4y+1\)

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

\(=3^2-4.3+1=-2\)

b)  \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

\(=7^2+2.7+37=100\)

c)  \(C=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)

\(=5^2-2.5+10=25\)

a) \(A=x^2+2xy+y^2-4x-4v+1\)

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

\(=3^2-4.3+1=-2\)

Đây nè bạn.

a) Ta có: \(M=x^2-2xy+y^2-10x+10y\)

\(=\left(x-y\right)^2-10\left(x-y\right)\)

\(=9^2-10\cdot9=-9\)

A= x²+2x +y²-2y-2xy+37

<=>A=(x²-2xy+y²)+(2x-2y)+37

<=>A=(x-y)²+2(x-y)+37=37²+2.37+37=1480

\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)

b: \(=7\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(7-x+y\right)=0\)

Lời giải:
$A=(x^2+2xy+y^2)+y^2-2\sqrt{2}(x+y)-2y+2022$

$=(x+y)^2-2\sqrt{2}(x+y)+2+(y^2-2y+1)+2019$

$=(x+y-\sqrt{2})^2+(y-1)^2+2019$

$\geq 2019$
Vậy $A_{\min}=2019$. Giá trị này đạt tại $x+y-\sqrt{2}=y-1=0$

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

a) Thay \(a =  - 4,b = 18\)vào đa thức ta có:

\(A =  - 5a - b - 20 =  - 5. - 4 - 18 - 20 =  - 18\).

b) Thay \(x =  - 1,y = 3,z =  - 2\)vào đa thức ta có:

\(B =  - 8xyz + 2xy + 16y =  - 8. - 1.3. - 2 + 2. - 1.3 + 16.3 =  - 48 - 6 + 48 =  - 6\).

c) Thay \(x =  - 2,y =  - 3\)vào đa thức ta có:

\(C =  - {x^{2021}}{y^2} + 9{x^{2021}} =  - {( - 1)^{2021}}.{( - 3)^2} + 9.{( - 1)^{2021}} =  - ( - 1).9 + 9.( - 1) = 9 + ( - 9) = 0\). 

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

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

\(=2\left(x^2+y^2\right)\ge\left(x+y\right)^2=4\)(BĐT Bunhiacopxki)

=> A \(\ge4\)Dấu "=" xảy ra <=> x=y=1

a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)

\(=-4.\dfrac{1}{4}+10=9\)

b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)

\(=\left(-2\right).\left(32-32\right)=0\)

a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)

\(=4x^2-4x+1+9-4x^2\)

\(=-4x+10\)

\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)