\(2x^2+xy+\dfrac{1}{4}y^2=2x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{8}y^2\)
\(=\dfrac{1}{2}x.\left(4x+y\right)+\dfrac{1}{8}y.\left(4x+y\right)\)
\(=\left(4x+y\right).\left(\dfrac{1}{2}x+\dfrac{1}{8}y\right)=\left(4x+y\right).\dfrac{1}{8}\left(4x+y\right)\)
\(=\dfrac{1}{8}\left(4x+y\right)^2\)
Vậy ô trống cần điền là:
\(2x^2+xy+...\dfrac{1}{4}y^2...=\dfrac{1}{8}\left(4x+y\right)^2\)
Câu b chuyển cộng thành trừ là xong!
\(2x^2+xy+\left(\dfrac{\sqrt{2}y}{4}\right)^2=\left(\sqrt{2}x+\dfrac{\sqrt{2}y}{4}\right)^2\)
\(\dfrac{x^2}{4}-xy+y^2=\left(\dfrac{x}{2}-y\right)^2\)
