uyaa,f đăng nhầm rr

Ta có : \(\left(x^4-2x^3+2x+a\right)\div\left(x-1\right)^2\)

\(\Rightarrow\left(x^4-2x^3+2x+a\right)\div\left(x^2-2x+1\right)\)

Ta đặt phép chia :

Để đa thức trên chia hết thì \(a+1=0\)

                                          \(\Rightarrow a=-1\)

Vậy \(a=-1\) thì \(\left(x^4-2x^3+2x+a\right)⋮\left(x-1\right)^2\)

thiếu dữ kiện a,b,c khác 0 nha

2(x + 1)(y + 1) = (x + y)(x + y + 2)

2(xy + y + x + 1) = x² + 2xy +y² + 2x + 2y

2xy + 2y + 2x + 2 = 2xy + 2y + 2x + x² + y²

2xy + 2y + 2x + x² + y² = 2xy + 2y + 2x + x² + y²

=> 2(x+1)(y+1)=(x+y)(x+y+2)

A=(n+1)4+n4+1=[(n2+2n+1)2−n2]+[(n4+2n2+1)−n2]A=(n+1)4+n4+1=[(n2+2n+1)2−n2]+[(n4+2n2+1)−n2]

=(n2+3n+1)(n2+n+1)+[(n2+1)2−n2]=(n2+3n+1)(n2+n+1)+[(n2+1)2−n2]

=(n2+3n+1)(n2+n+1)+(n2+n+1)(n2−n+1)=(n2+3n+1)(n2+n+1)+(n2+n+1)(n2−n+1)

=(n2+n+1)(n2+3n+1+n2−n+1)=(n2+n+1)(n2+3n+1+n2−n+1)

=(n2+n+1)(2n2+2n+1)=2.(n2+n+1)2⋮(n2+n+1)2=(n2+n+1)(2n2+2n+1)=2.(n2+n+1)2⋮(n2+n+1)2

⇒A⋮(n2+n+1)2⇒A⋮(n2+n+1)2 => đpcm

Chúc bạn học tốt

\(A=\left(n^2+2n+1\right)^2-n^2+\left(n^4+n^2+1\right)\)\(=\left(n^2+3n+1\right)\left(n^2+n+1\right)+\left(n^2+n+1\right)\left(n^2-n+1\right)\)

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