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

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

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

=2x(x-1) - 3(x-1)

=(x-1)(2x-3) xong

2x(x-1)-3x+3

=2x(x-1)-(3x-3)

=2x(x-1)-3(x-1)

=(x-1)(2x-3)

x³-3x²-3x+1=x³+1-3x²-3x

                  =(x+1)(x²-x+1)-3x(x+1)

                  =(x+1)(x²-x+1-3x)

                  =(x+1)(x²-4x+1)

x^3 - x + 3x^2y + 3xy^2 + y^3 - y

=x³+y³+3x²y+3xy²-x-y

=(x+y)(x²-xy+y²)+3xy(x+y)-(x+y)

=(x+y)(x²-xy+y²+3xy-1)

=(x+y)(x²+2xy+y²-1)

=(x+y)[(x+y)²-1]

=(x+y)(x+y-1)(x+y+1)

x^2 + 5x - 6

=x²-x+6x-6

=x.(x-1)+6.(x-1)

=(x-1)(x+6)

x^5+2x^4+2x^3+2x^2+2x+1

=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)

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

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

\(x^3-3x^2+3x-1\) =0

=>\(\left(x-1\right)^3\)=0

=>x-1=0

=>x=1

vậy x =1

\(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

x^3 + 3x -4 = x^3 - 1 + 3x - 3 = (x-1)(x^2-x+1) + 3(x-1) = (x-1)(x^2-x+1+3) = (x-1)(x^2-x+4)

nhớ cho mik nha

minh h cho 1  cai nhung o cho x^3-1 phai bang (x-1)(x^2+x+1)

ứng dụng t/c thêm bớt nà bạn