d/ $\left(2x-\dfrac{1}{2}\right)^2\\=(2x)^2-2.2x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\\=4x^2-2x+\dfrac{1}{4}$
Vậy $\left(2x-\dfrac{1}{2}\right)^2=4x^2-2x+\dfrac{1}{4}$
e/ $\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)\\=\left(\dfrac{x}{2}\right)^2-y^2\\=\dfrac{x^2}{4}-y^2$
Vậy $\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)=\dfrac{x^2}{4}-y^2$
