(x^2+x+1)*(x^5-x^4+x^2-x+1) 

( x^2 + x + 1 ) ( x^5 - x^4 + x^2 - x + 1 )

Ta có

x 7   –   x 2   –   1   =   x 7   –   x   –   x 2   +   x   –   1     =   x ( x 6   –   1 )   –   ( x 2   –   x   +   1 )     =   x ( x 3   –   1 ) ( x 3   +   1 )   –   ( x 2   –   x   +   1 )     =   x ( x 3   –   1 ) ( x   +   1 ) ( x 2   –   x   +   1 )   –   ( x 2   –   x   +   1 )     =   ( x 2   –   x   +   1 ) [ x ( x 3   –   1 ) ( x   +   1 )   –   1 ]     = x 2 − x + 1 x 4 − x x + 1 − 1 = x 2 − x + 1 x 5 + x 4 − x 2 − x − 1

Đáp án cần chọn là: B

-Đặt \(t=\left(x^2-x+1\right)\)

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

\(=t^2-5xt+4x^2\)

\(=t^2-4xt-xt+4x^2\)

\(=t\left(t-4x\right)-x\left(t-4x\right)\)

\(=\left(t-4x\right)\left(t-x\right)\)

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

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

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

\(\left(x^2+x+1\right)\left(x^2+x+5\right)-21=x^4+x^3+5x^2+x^3+x^2+5x+x^2+x+5-21=x^4+2x^3+7x^2+6x-16=\left(x-1\right)\left(x+2\right)\left(x^2+x+8\right)\)

\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-21\)

\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-21\)

\(=\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+7\left(x^2+x+1\right)-21\)

\(=\left(x^2+x+1\right)\left(x^2+x-2\right)+7\left(x^2+x-2\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x+8\right)\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2+x+8\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+5\right)-21\)

\(=\left(x^2+x\right)^2+6\left(x^2+x\right)+5-21\)

\(=\left(x^2+x\right)^2+6\left(x^2+x\right)-16\)

\(=\left(x^2+x+8\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+8\right)\left(x+2\right)\left(x-1\right)\)

a,

\(A=4(x-2)(x+1)+(2x-4)^2+(x+1)^2\\=[2(x-2)]^2+2\cdot2(x-2)(x+1)+(x+1)^2\\=[2(x-2)+(x+1)]^2\\=(2x-4+x+1)^2\\=(3x-3)^2\)

Thay $x=\dfrac12$ vào $A$, ta được:

\(A=\Bigg(3\cdot\dfrac12-3\Bigg)^2=\Bigg(\dfrac{-3}{2}\Bigg)^2=\dfrac94\)

Vậy $A=\dfrac94$ khi $x=\dfrac12$.

b,

\(B=x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\=(x^9-1)-(x^7-x^4)-(x^6-x^3)-(x^5-x^2)\\=[(x^3)^3-1]-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1)-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1-x^4-x^3-x^2)\\=(x^3-1)(x^6-x^4-x^2+1)\)

Thay $x=1$ vào $B$, ta được:

\(B=(1^3-1)(1^6-1^4-1^2+1)=0\)

Vậy $B=0$ khi $x=1$.

$Toru$

Là nhân tử rồi bn ơi

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

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

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

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

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

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

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

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

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