minh hiền làm sai rồi

bạn hoàng tính như nào mà ra đc dòng cuối vậy?

\(x^2+4y^2+4xy-10x-20y+25\)

\(=\left(x^2-10x+25\right)+\left(4xy-20y\right)+4y^2\)

\(=\left(x-5\right)^2+4y\left(x-5\right)+4y^2\)

\(=\left(x-5\right)\left(x-5+4y\right)+4y^2\)

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

(x+y)^2 - 5y^2

x^2- 4y^2 + 4xy

= x^2 + 4xy - 4y^2

=x^2 + 2x2y - (2y)^2

= ( x - 2y )^2

 x² - 4y² + 4xy = x² + 4xy - 4y² = - (x² - 4xy + 4y²) = - ((x² - 2x.2y + (2y)²)

= - ( x - 2y)²

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

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

\(\left(3-5x\right)^2\)

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

\(-x^2+6x-9=-\left(x-3\right)^2\)

\(25x^2-30x+9=\left(5x-3\right)^2\)

x²-4xy+4y²-x+2y=(x-2y)²-(x-2y)

                         =(x-2y)(x-2y-1)