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

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

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

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

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

a,\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)=3x^3y^3-x^2y^2+\frac{3}{5}x^3y^2\)

b,\(5x^3-5x=5x\left(x^2-1\right)=5x\left(x-1\right)\left(x+1\right)\)

c, \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(3x-5\right)\left(x-y\right)\)

1) 1/5x²y( 15xy² - 5y + 3xy ) = 3x³y³ - x²y² + 3/5x³y²

2) a) 5x³ - 5x = 5x( x² - 1 ) = 5x( x² - 1² ) = 5x( x - 1 )( x + 1 )

b) 3x² + 5y - 3xy - 5x = ( 3x² - 3xy ) + ( 5y - 5x )

                                  = 3x( x - y ) + 5( y - x )

                                  = 3x( x - y ) + 5[ -( x - y ) ]

                                  = 3x( x - y ) - 5( x - y )

                                  = ( 3x - 5 )( x - y )

Bài 1 :

\(15x^2y\left(15xy^2-5y+3xy\right)=3x^3y^3-x^2y^2+\frac{3}{5}x^3y^2\)

Bài 2 : 

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

giúp mk

9² - 30xy + 25y²

= ( 3x )² - 2 . 3x . 5y + ( 5y )²

= ( 3x - 5y )²

Sai thì thôi

thank

ko sao

Biểu thức này không phân tích được thành nhân tử nhé. Bạn xem xem có viết sai đề không.

\(5x^3-5x=5x\left(x^2-1\right)\)

\(3x^2+5x-3xy-5x=x\left(3x+5\right)-x\left(3y+5\right)=x\left(3x-3y\right)=3x\left(x-y\right)\)

\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)\)

\(=\frac{1}{5}x^2y^2\left(15xy-5+3x\right)\)

\(=\frac{1}{5}\left(x.y\right)^2.\left(15xy-5+3x\right)\)

\(=\frac{1}{5}\left(15x^3y^3-5x^2y^2+3x^3y^2\right)\)

làm phần 1 đi bạn

\(1,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\\ 2,4x^2-4x+1-25y^2=\left(2x-1\right)^2-\left(5y\right)^2=\left(2x-5y-1\right)\left(2x+5y-1\right)\\ c,x^2-3x-10=\left(x^2-5x\right)+\left(2x-10\right)=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+2\right)\)

ab hay là xy bạn ơi