Question

\(2x^{^{ }2}+xy+\) ______= (______+_____)\(^2\)

______ \(-xy+y^2=\) ( ______\(-\) _____ )\(^2\)

Ai chơi piano bài Bluestone Alley k ag ???

Answer 1

\(2x^2+xy+\dfrac{1}{4}y^2=2x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{8}y^2\)

\(=\dfrac{1}{2}x.\left(4x+y\right)+\dfrac{1}{8}y.\left(4x+y\right)\)

\(=\left(4x+y\right).\left(\dfrac{1}{2}x+\dfrac{1}{8}y\right)=\left(4x+y\right).\dfrac{1}{8}\left(4x+y\right)\)

\(=\dfrac{1}{8}\left(4x+y\right)^2\)

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!

Answer 2

\(2x^2+xy+\left(\dfrac{\sqrt{2}y}{4}\right)^2=\left(\sqrt{2}x+\dfrac{\sqrt{2}y}{4}\right)^2\)

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

Answer 3

cái hell gì đây