A   =   x 2   +   2 x y   +   y 2   –   4 x   –   4 y   +   1     =   ( x 2   +   2 x y   +   y 2 )   –   ( 4 x   +   4 y )   +   1     =   ( x   +   y ) 2   –   4 ( x   +   y )   +   1

Tại x + y = 3, ta có: A = 3 2 – 4.3 + 1 = -2

Đáp án cần chọn là: D

a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)

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

\(=3x^2+3y^2=3\)

b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)

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

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

=9-12+1

=-2

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

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

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

\(=\left(x-y\right)^2+2\left(x-y\right)=7^2+2\cdot7=49+14=63\)

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

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

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

\(=7^3-7^2=343-49=294\)

b: \(C=x^2+4y^2-2x+10+4xy-4y\)

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

\(=\left(x+2y\right)^2-2\left(x+2y\right)+10=5^2-2\cdot5+10=25\)

Sửa đề: \(a=x^2+2xy+y^2-4x-4y-2\)

Ta có: \(a=x^2+2xy+y^2-4x-4y-2\)

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

\(=3^2-4\cdot3-2\)

=9-12-2

=-3-2

=-5

\(A=4\cdot3\left(-2\right)-2\left(3+2\right)=-24-10=-34\\ B=\left(x+y\right)^2-3\left(x+y\right)=\left(x+y\right)\left(x+y-3\right)=\left(x+y\right)\left(2+1-3\right)=0\)

Bài 3: 

a) Ta có: \(A=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)

d) Ta có: \(D=x^2-2x+2\)

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

\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)

Bài 1: 

a) Ta có: \(A=x^2-2x+5\)

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

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1

b) Ta có: \(B=x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)