bạn viết lại đề đc ko bạn:>,ko hỉu đề

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Cộng à bn

\(S=2^{2019}-2^{2018}-2^{2017}-...-2^2-2-1\)

   \(=2^{2019}-\left(1+2+2^2+...+2^{2017}+2^{2018}\right)\) (1)

Đặt \(Q=1+2+2^2+...+2^{2017}+2^{2018}\)

\(2Q=2+2^2+2^3+...+2^{2018}+2^{2019}\)

\(2Q-Q=2^{2019}-1\)

\(Q=2^{2019}-1\)(2) 

Từ (1) và (2), ta được:

\(S=2^{2019}-\left(2^{2019}-1\right)=1\)

\(S=2+2^2+...+2^{2016}\)

\(\Rightarrow2S=2.\left(2+2^2+...+2^{2016}\right)\)

\(\Rightarrow2S=2^2+2^3+...+2^{2017}\)

\(\Rightarrow2S-S=2^2+2^3+...+2^{2017}-\left(2+2^2+...+2^{2016}\right)\)

\(\Rightarrow S=2^2+2^3+...+2^{2017}-2-2^2-...-2^{2016}\)

\(\Rightarrow S=2^{2017-2}\)

2^1+2^2+2^3+2^4+2^2016

=2^1+2+3+4+2016

=2^2026

S=2 + 2² + 2³+...+2²⁰¹⁶

2.S= 2²+2³+2⁴+......+2²⁰¹⁷

2.S - S = (2²+2³+2⁴+......+2²⁰¹⁷) - (2 + 2² + 2³+...+2²⁰¹⁶)

S = 2²⁰¹⁷-2

Bài 1a

B=4/1.3 + 4/3.5 + 4/5.7+...+4/2017.2019

B=4.2/(1.3).2 + 4.2/(3.5).2 + 4.2/(5.7).2+....+4.2/(2017.2019).2

B=2.( 2/1.3 + 2/3.5 + 2/5.7 +...+ 2/2017.2019 )

B=2.(1-1/3+1/3-1/5+1/5-1/7+....+1/2017-1/2019)

B=2.(1-1/2019)

B=2.(2019/2019-1/2019)

B=2.2018/2019

B=4036/2019

cảm ơn

Câu B bài 1 là 2017 + 2018 hay 2017 x 2018 vậy bạn

à mik ghi nhầm , là 2017 nhân 2019 nha bn 

B1:

B = 2.2/1.3 + 2. 2/3.5 + 2.2/5.7 + .... + 2.2/2017.2019

B= 2. ( 2/1.3 + 2/3.5 + 2/5.7 +....+ 2/2017.2019)

B = 2 . ( 1-1/3+1/3-1/5+1/5-1/7+...+1/2017-1/2019)

B= 2. ( 1-1/2019)

B=2.2018/2019

B=4036/2019

\(2^0+2^1+2^2+2^3+2^0.2^1.2^2.2^3\)3

\(=1+2+4+8+1.2.4.8\)

\(=15+64=79\)

n = 2⁵.3^{x^2}

Vì n có 60 ước nên ta có:

(5 + 1)(x² + 1) = 60

=> 6(x² + 1) = 60

=> x² + 1 = 10

=> x² = 9

=> x = 3 (Vì x thuộc N*)

Do đó n = 2⁵.3^{3^2} = 2⁵.3⁹ = 629856

629856 , ủng hộ mk nha

n=2592 mới đúng chứ