x^2-9=x^2-3^2

         =(x-3).(x+3)

`x^2-9=x^2 - 3^2=(x-3)(x+3)`

\(^{x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)}\)

x^2-x-6=x^2-3x+2x-6

            =(x^2-3x)+(2x-6)

            =x(x-3)+2(x-3)

            =(x+2).(x-3)

`-2x +x=A.x => A=(-2x+x)/x =>A=-2 +1=-1`

\(n^5-n=n\left(n^4-1\right)=n\left(n^2-1\right)\left(n^2+1\right)\)

                                             =\(n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)

                                            \(=n\left(n-1\right)\left(n+1\right)\left[\left(n^2-4\right)+5\right]\)

                                             \(=n\left(n-1\right)\left(n+1\right)\left(n-2\right)\left(n+2\right)-5n\left(n-1\right)\left(n+1\right)\)

                                              \(=\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)-5n\left(n-1\right)\left(n+1\right)\)

Do \(\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)là 5 số nguyên dương liên tiếp \(\Rightarrow\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮5\)(1)

Do    \(5⋮5\Rightarrow5n\left(n-1\right)\left(n+1\right)⋮5\)(2)

Từ (1) và (2) => ĐPCM

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

\(\left(x-3+2-x\right)\left(x^2-3x+9-2x+x^2+6-3x+x^2-2x+4\right)\)

\(-\left(3x^2-10x+19\right)\)

\(10x-3x^2-19\)

x [x - 1] + x [ x + 3 ]

= x [ x - 1 + x + 3 ]

= x [ 2 x + 2 ]

= 2 x [ x + 2 ]

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

?????????

\(x^4-2x^3+x^2-2x\)

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

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

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