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) \(A=x^2+2xy+y^2-4x-4v+1\)

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

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

=25 nha nhớ tích

A=(x^2+4xy+4y^2)-(2x+4y)+10

A=(x+2y)^2-2(x+2y)+1+9

A=(x+2y-1)^2+9

A=(5-1)^2+9=16+9=25

A=(x^2+4xy+4y^2)-(2x+4y)+10

A=(x+2y)^2-2(x+2y)+1+9

A=(x+2y-1)^2+9

A=(5-1)^2+9=16+9=25

a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

(x²+4xy+4y²)-(2x+4y)+10=(x+2y)²-2(x+2y)+10=5²-10+10=25 :333

:333 là biểu cảm nhé

​

Theo đề ta có : x + 2y = 5

và A = \(x^2-4y^2-2x+10+4xy-4y\)

\(=\left(x^2-4y^2+4xy\right)\) - \(\left(2x+4y\right)+10\)

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

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

=25

k nha mn!

TA có A=\(A=x^2-4y^2-2x+10+4xy-4xy\)

\(=\left(x^2-4y^2=4xy\right)-\left(2x+4y\right)+10\)

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

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

\(=25\)

tích đúng mình làm cho

là sao ạk
giải giùm mình với ạk

1/Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

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

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

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

Thay x-y=7 vào A ta đc:

\(A=7^2+2.7+37=49+14+37=100\)

2/Ta có :\(B=x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

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

Thay x+2y=5 vào B ta đc

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

x² + 4y² - 2x + 10 + 4xy - 4y

= (x² + 4xy + 4y²) - 2x + 10 - 4y)

= (x + 2y)² - (2x + 4y) + 10

= 5² - 2(x + 2y) + 10

= 25 - 10 + 10

= 25

\(x^2+4y^2-2x+10+4xy-4y\) =\(x^2+4xy+4y^2-2\left(x+2y\right)+10\)

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

                                                                         =\(5^2-2\cdot5+10=25\)

\(x^2+4y^2-2x+10+4xy-4y\)

\(=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)

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

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

\(=25\)

B=\(x^2+4y^2-2x+10+4xy-4y\)

B=\(x^2+4xy+4y^2-2\left(x+2y\right)+10\)

B=\(\left(x+2y\right)^2-2\left(5\right)+10\)

B=\(5^2-10+10\)

B=25

\(B=x^2+4y^2-2x+10+4xy-4y\)

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

\(B=x^2+4y^2+4xy\)

\(B=x^2+2.x.2y+\left(2y\right)^2=\left(x+2y\right)^2\)

\(B=5^2\)

\(B=25\)