\(\left(2x^2-4x-\frac{1}{4}y\right)^2\)
= \(\left(2x^2-4x-\frac{1}{4}y\right)\left(2x^2-4x-\frac{1}{4}y\right)\)
= \(4x^4-8x^3-\frac{1}{2}x^2y-8x^3+16x^2+xy-\frac{1}{2}x^2y+xy+\frac{1}{16}y^2\)
= \(4x^4-16x^3-x^2y+16x^2+2xy+\frac{1}{16}y^2\)
chắc s rùi :((