Ta có: A = x2 - 4y2 + 4y = x2 - 4y(y - 1) = x2 - 4y[y - (x + 2y)] = x2 - 4y(y - x - 2y) = x2 - 4y(-y - x) = x2 + 4y2 + 4xy = (x + 2y)2 = 12 = 1
Ta có :
\(A=x^2-4y^2+4y\)
\(=\left(x^2-4y^2\right)+4y\)
\(=\left(x-2y\right)\left(x+2y\right)+4y\)
\(=x-2y+4y\) \(\left(x+2y=1\right)\)
\(=x+2y\)
\(=1\)
Vậy...........
Học tốt!!!
A= x2 -4y2+4y
=x(x-2y)-2xy+4y2+4y
= x+2y-4y2-2xy+2y
=1-2y(2y+x-1)
=1-2y(1-1)
=1-0=1