a) \(13\times17-256:16+14:7-1\)

\(=221-16+2-1\)

\(=206\)

b: \(B=2\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\cdot\left(2+...+2^{59}\right)⋮3\)

\(B=2+2^2+...+2^{60}\)

\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{58}\right)⋮7\)

2004¹⁰+2004¹⁰>2005¹⁰

a) Lập bảng

Ta có: 2018 : 4 = 504 (dư 2)

Suy ra \(2017^{2018}+2019^{2018}= \overline{...9}+\overline{...1}=\overline{...0}\)

Vậy 2017²⁰¹⁸ + 2019²⁰¹⁸ chia hết cho 10

b) Làm tương tự như câu a)

\(\left(\dfrac{2003}{2004}\right)^0-\left(\dfrac{1}{3}\right)^3\div\left(\dfrac{1}{3}\right)^2\)

\(=1-\dfrac{1}{3}\)

\(=\dfrac{2}{3}\)

A=1-1/3=2/3

Ta có:

A= 2+2²+2³+…+2²⁰⁰⁴

A=2(1+2)+2³(1+2)+…+2²⁰⁰³(1+2)

Vậy A chia hết cho 3.

A=2(1+2+2²) + 2⁴(1+2+2²)+…+2²⁰⁰²(1+2+2²).

Vậy A chia hết cho 7.

A=2(1+2+2²+2³)+2⁵(1+2+2²+2³)+…+2²⁰⁰¹ (1+2+2²+2³)

Vậy A chia hết cho 15.

thôi cả 2 bạn k nên bực bội với nhau làm j cho mất công tốn tg

A chia het cho 15 

x=2005

nên x+1=2006

\(f\left(x\right)=x^{2005}-x^{2004}\left(x+1\right)+x^3\left(x+1\right)-...+x\left(x+1\right)\)

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

=x=2005