a: \(3x^3y^2-12x^2y^3\)

\(=3x^2y^2\cdot x-3x^2y^2\cdot4y\)

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

b: \(x^2-5x+xy-5y\)

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

=(x-5)(x+y)

d: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

\(=2\left\lbrack\left(x-y\right)^2-3\right\rbrack\)

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

e: \(x^3-2020x_{}^2+2020x-2019\)

\(=x^3-x^2+x-2019x^2+2019x-2019\)

\(=x\left(x^2-x+1\right)-2019\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(x-2019\right)\)

1+1=mấy

1+1=2 chứ bao nhiêu

2020X4-2000X4=0

= ( 2020 + 2020 ) + ( 2020 + 2020 ) - ( 2020 x 4 )

=         4040         +       4040            -      8080
=                      8080            -            8080
=                                   0

\(x^{2019}-2020x^{2018}+2020x^{2017}-2020x^{2016}+...+2020x-2020\)

\(=x^{2019}-2019x^{2018}-x^{2018}+2019x^{2017}+x^{2017}\)

\(-2019x^{2016}-x^{2016}+...+2019x+x-2020\)

\(=x^{2018}\left(x-2019\right)-x^{2017}\left(x-2019\right)+x^{2016}\left(x-2019\right)\)

\(+...-x\left(x-2019\right)+\left(x-2019\right)-1\)

\(=-1\)

a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

\(=\left(x+1\right)^2\)

\(=x^2+2x+1\)