\(A=x^6-2007x^5+2007x^4-2007x^3+2007x^2-2007x+2007\)

\(=x^6-2006x^5-x^5+2006x^4+x^4-2006x^3-x^3+2006x^2+x^2-2006x-x+2006+1\)

\(=x^5\left(x-2006\right)-x^4\left(x-2006\right)+x^3\left(x-2006\right)-x^2\left(x-2006\right)+x\left(x-2006\right)-\left(x-2006\right)+1\)

\(=\left(x^5-x^4+x^3-x^2+x-1\right)\left(x-2006\right)+1\)

Thay x = 2006

\(\Leftrightarrow A=1\)

Vậy A = 1 tại x = 2006

\(A=x^6-2007.x^5+2007.x^4-2007.x^3+2007.x^2-2007.x+2007\)

\(=x^6-\left(x+1\right).x^5+\left(x+1\right).x^4-...+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-x^4-...-x+1\)

\(=1\)

2007+2007+2007+2007 x 4

= 2007 x 3+2007 x 4

= 2007 x (3+4)

= 2007 x 7

= 14049

\(2007+2007+2007+2007\times4\)

\(=2007\times3+2007\times4\)

\(=2007\times\left(3+4\right)\)

\(=2007\times7\)

\(=14049\)

Ta có: \(\hept{\begin{cases}x=\frac{y}{2}\\\frac{y}{3}=\frac{z}{4}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{3}=\frac{y}{6}\\\frac{y}{6}=\frac{z}{8}\end{cases}}\Rightarrow\frac{x}{3}=\frac{y}{6}=\frac{z}{8}\)

Đặt: \(\frac{x}{3}=\frac{y}{6}=\frac{z}{8}=k\Rightarrow\hept{\begin{cases}x=3k\\y=6k\\z=8k\end{cases}}\)

Khi đó \(\frac{x+y+z}{x+y-z}=\frac{3k+6k+8k}{3k+6k-8k}=17\)

b) Từ \(ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\frac{a^{2017}}{c^{2017}}=\frac{b^{2017}}{d^{2017}}=\left(\frac{a-b}{c-d}\right)^{2017}\)(1)

Mặt khác: \(\frac{a^{2017}}{c^{2017}}=\frac{b^{2017}}{d^{2017}}=\frac{a^{2017}-b^{2017}}{c^{2017}-d^{2017}}\)(2)

Từ (1) và (2) =>đpcm

cảm ơn girl nhưng phần b là mũ 2007 bạn nhé

https://hentaiz.net/

Ta co

A=2007^2006( lên lơp 6 e se hoc)

=>A=2007^2 x 2007^2004

=>(...9)x(...1)=(...9)     (1)

Ta co:

B=2006^2007=(...6)

A=....9

B=....6

A-B= ....9- ....6= ....3 không chia hết cho 5