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

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

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

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

\(=x\)

\(=4\)

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

5 x 3 + 5 x 4 + 5 x 2 + 5

= 5 x 3 + 5 x 4 + 5 x 2 + 5 x 1

= 5 x ( 3 + 4 + 2 + 1 )

= 5 x 10

= 50

  5 x 3 + 5 x 4 + 5 x 2 + 5

 = 5 x (3 + 4 + 2) + 5

= 5 x 9 + 5

= 45 + 5

= 50

=5x(3+4+2+1)

=5x10

=50

b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)

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

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

c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=5x^3+14x^2+12x+8\)

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

1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)

\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)

\(=5x^3+14x^2+12x+8\)

d) Ta có: \(\dfrac{5x^3+14x^2+12x+8}{x+2}\)

\(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}\)

\(=\dfrac{5x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)}{x+2}\)

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

c) Ta có: \(\dfrac{5x^4+9x^3-2x^2-4x-8}{x-1}\)

\(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

\(=\dfrac{5x^3\left(x-1\right)+14x^2\left(x-1\right)+12x\left(x-1\right)+8\left(x-1\right)}{x-1}\)

\(=5x^3+14x^2+12x+8\)

4 + 4 x 2 + 4 x 3 + 4 x 4 = ( 4 x 2 + 4 x 3 ) + (4 x 4 + 4) 

                                      =   20 + 20 

                                        = 40 

5 + 5 x 2 + 5 x 3 + 5 x 4 = (5 x 4 + 5 x 2 ) + ( 5 x 3 + 5 )

                                      = 30 + 20 

                                     = 50