D

\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(1+...+7^{118}\right)⋮57\)

3^x x 2 + 15 = 33

<=> 3^x = 9

<=> x = 2

3^x . 2 + 15 = 33

3^x . 2 = 33 - 15 = 18

3^x = 18 : 2 = 9 = 3²

=> x = 2

Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(A=\dfrac{1}{2}-0-0-0-...-0-\dfrac{1}{100}\)

\(A=\dfrac{1}{2}-\dfrac{1}{100}\)

\(A=\dfrac{50}{100}-\dfrac{1}{100}\)

\(A=\dfrac{49}{100}\)

c: Trường hợp 1: n=2k

\(\Leftrightarrow n\left(n+2017\right)=2k\left(2k+2017\right)⋮2\)

Trường hợp 2: n=2k+1

\(\Leftrightarrow n\left(n+2017\right)=\left(2k+1\right)\left(2k+2018\right)⋮2\)

2S = 2 + 2^2 + 2^3 + ...+ 2^64

2S + 1 = 1 + 2 + 2^2 + ... + 2^64

2S - S = 2^64 - 1

Vậy S =  2^64 - 1

\(A=2^2+2^3+...+2^{62}+2^{63}\)

\(2A=2^3+2^4+...+2^{63}+2^{64}\)

\(2A-A=\left(2^3+2^4+...+2^{63}+2^{64}\right)-\left(2^2+2^3+...+2^{62}+2^{63}\right)\)

\(A=2^{64}-2^2\)

\(a,\Rightarrow20\cdot2^x=160+1-1\\ \Rightarrow2^x=160:20=8=2^3\\ \Rightarrow x=3\\ b,\Rightarrow\left(4-x:2\right)^3=2\left(8-5\right)+1+1\\ \Rightarrow\left(4-x:2\right)^3=6+2=8=2^3\\ \Rightarrow4-x:2=2\\ \Rightarrow x:2=2\Rightarrow x=4\\ c,n\left(n+2017\right)\)

Với n chẵn thì \(n=2k\left(k\in N\right)\Rightarrow n\left(n+2017\right)=2k\left(n+2017\right)⋮2\)

Với n lẻ thì \(n=2k+1\left(k\in N\right)\Rightarrow n\left(n+2017\right)=n\left(2k+2018\right)=2n\left(k+1009\right)⋮2\)

Vậy \(n\left(n+2017\right)\) luôn chẵn

\(d,3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}=\left(2^3\right)^{100}=2^{300}\)

chịu

hỏi thế ai trả lời dc