\(x=7\Rightarrow8=x+1\left(1\right)\)

Thay \(1\) vào \(F\) ta có:

\(F=x^{2006}-\left(x+1\right)^{2005}+\left(x+1\right)^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\)

\(F=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\)

\(F=-7-5\)

\(\Rightarrow F=-12\)

x=4

=>x+1=5

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

  =x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+1

  =x^6-x-1

  =4^6-4-1

  =4091

\(a,A=5\cdot4^5-5\cdot4^4+5\cdot4^3-5\cdot4^2+5\cdot4+1\\ A=4^4\left(20-5\right)+4^2\left(20-5\right)+\left(20-5\right)\\ A=15\left(4^4+4^2+1\right)=15\cdot273=4095\)

\(b,x=7\Leftrightarrow x+1=8\\ \Leftrightarrow B=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\\ B=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\\ B=-x-5=-12\)

b)tương tự

=x^2006-x^2006-x^2005+x^2005+x^2004-...+x^3-x^2-x^2-x-5

=-x-5

=-7-5=-12

Thay x = 4 vào A ta được:

5.4⁵ - 5.4⁴ + 5.4³ - 5.4² + 5.4 - 1

= 5.1024 - 5.256 + 5.64 - 5.16 + 5.4 - 1

= 5120 - 1280 + 320 - 80 + 20 - 1

= 4099

Kq = 4099 nha

bằng 10

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

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

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

\(=x\)

\(=4\)

a) 

\(P=\left(x^{14}-9x^{13}\right)-\left(x^{13}-9x^{12}\right)+\left(x^{12}-9x^{11}\right)-...+\left(x^2-9x\right)-\left(x-9\right)+1\)

\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+x^{11}\left(x-9\right)+...+x\left(x-9\right)-\left(x-9\right)+1\)

\(P\left(9\right)=1\)

b)

\(Q=\left(x^{15}-7x^{14}\right)-\left(x^{14}-7x^{13}\right)+\left(x^{13}-7x^{12}\right)-...-\left(x^2-7x\right)+\left(x-7\right)+2\)

\(=x^{14}\left(x-7\right)-x^{13}\left(x-7\right)+x^{12}\left(x-7\right)-...-x\left(x-7\right)+\left(x-7\right)+2\)

\(Q\left(7\right)=2\)

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

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

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

\(=-2\)

1)Ta có:x=4=>x+1=5(1)

Mặt khác:A=x⁵-5x⁴+5x³-5x²+5x-1(2)

Thay (1) vào (2) ta có:

A=x⁵-(x+1)x⁴+(x+1)x³-(x+1)x²+(x+1)x-1

=>A=x⁵-x⁵-x⁴+x⁴+x³-x³-x²+x²+x-1

=>A=x-1=4-1=3

2)Vì a:5 dư 2,b:5 dư 3 nên:

Đặt:a=5x+2;b=5y+3

Khi đó:ab=(5x+2)(5y+3)=25xy+10y+15x+6

=5(5xy+2y+3x+1)+1

Vì 5(5xy+2y+3x+1)\(⋮\)5 nên =>5(5xy+2y+3x+1)+1:5 dư 1 hay ab:5 dư 1

Vậy ab:5 dư 1

3)

a)Nhận xét:

a₁=1

a₂=1+2=3

a₃=1+2+3=6

a₄=1+2+3+4=10

Khi đó:a₁₀₀=1+2+3+...+100=\(\dfrac{100.101}{2}\)=5050

a_n=1+2+3+...+n=\(\dfrac{n\left(n+1\right)}{2}\)

b)Gọi 2 số hạng liên tiếp là n-1;n

=>a_n-1=1+2+3+...+(n-1)=\(\dfrac{\left(n-1\right)n}{2}\)

=>a_n=\(\dfrac{\left(n+1\right)n}{2}\)(ở câu a)

Khi đó:tổng 2 số hạng liên tiếp là a_n+a_n-1 là:

a_n+a_n-1=\(\dfrac{n\left(n+1\right)+n\left(n-1\right)}{2}\)=\(\dfrac{2n.n}{2}\)

=\(\dfrac{2n^2}{2}\)=n² là số chính phương

Vậy tổng 2 số hạng liên tiếp là số chính phương

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

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

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

\(=3\)

Ta có : 

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

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

\(A=3\)

P/s tham khảo nha 

hok tốt

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

\(\text{Thay x=-1 vào biểu thức A,ta được:}\)

\(A=3.\left(-1\right)^5-3.\left(-1\right)^4+5.\left(-1\right)^3-\left(-1\right)^2+5.\left(-1\right)+2\)

\(A=3.\left(-1\right)-3.1+5.\left(-1\right)-1+5.\left(-1\right)+2\)

\(A=\left(-3\right)-3+\left(-5\right)-1+\left(-5\right)+2\)

\(A=\left(-6\right)+\left(-5\right)-1+\left(-5\right)+2\)

\(A=\left(-11\right)-1+\left(-5\right)+2\)

\(A=\left(-12\right)+\left(-5\right)+2\)

\(A=\left(-17\right)+2=-15\)

Thay x=-1 vào A ta có:
A= 3x⁵-3x⁴+5x³-x²+5x+2

   = 3.(-1)⁵-3.(-1)⁴+5.(-1)³-(-1)²+5.(-1)+2

    = 3.(-1)-3.1+5.(-1)-1+(-5)+2

    = -3-3-5-1-5+2

    =-15