Question

[Min Trâm]

Phân tích đa thức sau thành nhân tử

a ) 9(x+y-1)^2 - 4 (2x+3y+1)^2

b ) 3x^4y^2 +3x^3y^2 +3xy^2 +3y^2

c ) ( x+y )^3 - 1 -3xy( x + y -1)

d ) x^3 + 3x^2 + 3x +1 - 27z^3

Giúp với ạ ! Cảm ơn

Answer 1

[Minh Anh]

a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)

\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)

\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)

\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)

\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)

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

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

c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)

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