\(V\text{ì}\) \(tamgi\text{ác}\) ABC \(c\text{â}n\) \(t\text{ại}\) B

⇒BA=BC 

\(M\text{à}\) BE=BF

⇒EA=CF

⇒\(T\text{ứ}\) \(gi\text{ác}\) AEFC \(l\text{à}\) \(h\text{ình}\) \(thang\) \(c\text{â}n\)

=9\(x^4\)+4xy+\(\dfrac{4}{9}y^2\)

mik viết thiếu sorry:)))

(3x²+\(\dfrac{2}{3}\)y)²

=9\(x^4\)+\(\dfrac{4}{9}\)\(y^2\)

A=9x^2−6x+1

=(3x)^2−2.3x.1+1^2

=(3x−1)^2

B=(2x+3y)^2+(2x+3y)+1(2x+3y)2+(2x+3y)+1

=[(2x+3y)^2+2.(2x+3y).1/2+(1/2)^2]+3/4

=(2x+3y+1/2)^2+3/4

=(2x+3y+1)(2x+3y)+1