ab hay là xy bạn ơi

ab hay xy bạn

cả hai ab và xy

a) \(=x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)

b) \(=x^2-\left(3y\right)^2=\left(x-3y\right)\left(x+3y\right)\)

c) \(=\left(2x-1\right)^2-\left(2y\right)^2=\left(2x-1-2y\right)\left(2x-1+2y\right)\)

d) \(=x^2-10xy+\left(5y\right)^2=\left(x-5y\right)^2\)

e) \(=\left(3x\right)^2-6x+1=\left(3x-1\right)^2\)

f) \(=\left(5x\right)^2+20x+4=\left(5x+2\right)^2\)

\(a)x^2-4y^2=(x-2y)(x+2y)\\b)x^2-9y^2=(x-3y)(x+3y)\\c)(2x-1)^2-4y^2=(2x-1-2y)(2x-1+2y)\\d) x^2-10xy+25y^2=(x-5y)^2\\e)9x^2-6x+1=(3x-1)^2\\f)25x^2+20x+4=(5x+2)^2\)

\(x^2+3cd\left(2-3cd\right)-10xy-1+25y^2=x^2+6cd-\left(3cd\right)^2-10xy-1+\left(5y\right)^2\\ \)

\(=x^2-10xy+\left(5y\right)^2-\left(1-6cd+\left(3cd\right)^2\right)\)

\(=\left(x-5y\right)^2+6cd-1-\left(3cd\right)^2=\left(x-5y\right)^2-\left(1-3cd\right)^2\)

\(=\left(x-5y-1+3cd\right)\left(x-5y+1-3cd\right)\)

(x^2-10xy+25y^2)-(1-3cd(2-3cd))

=(x^2-2.x.5.y+(5y)²)-(1-6cd+9(cd)²)

=(x-5)^2-((3cd)²-2.3cd.1+1²)

=(x-5)^2-(3cd-1)²

=(x-5+3cd-1).(x-5-3cd+1)

=(x+3cd-6).(x-3cd-4)...

Theo mk là z.  Có thể phân tích nữa,đúng hay sai mk cx chưa chắc chắn vì mk cũng mới học thui.

\(x^2+2xy+y^2-25y^2\)

\(=\left(x^2+2xy+y^2\right)-25y^2\)

\(=\left(x+y\right)^2-\left(5y\right)^2\)

\(=\left(x+y-5y\right).\left(x+y+5y\right)\)

\(=\left(x-4y\right).\left(x+6y\right)\)