\(A=x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1=3^2-12+1=-2\)
\(B=x^2-2xy+y^2-5x+5y+6=\left(x-y\right)^2-5\left(x-y\right)+6=7^2-5.7+6=20\)
a)Ta có
A=\(x^2+2xy+y^2-4x-4y+1\)
=>A=\(\left(x+y\right)^2-4\left(x+y\right)+1\)
Mà x+y=3 nên
A=\(3^2-4\cdot3+1\)
A=-2
b)Ta có:
B=\(x^2-2xy+y^2-5x+5y+6\)
B=\(\left(x-y\right)^2-5\left(x-y\right)+6\)
Mà x-y=7 nên
B=\(7^2-5\cdot7+6\)
B=20