Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)

\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)

\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)

\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)

Giờ ta thế vào bài toán ban đầu được

\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)

\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)

\(=\frac{2017}{2018}\)  

a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)

Đặt \(B=2+2^2+...+2^{2017}\)

\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)

\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)

\(\Rightarrow B=2^{2018}-2\)

\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)

\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)

\(\Rightarrow A=-2\)

b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)

 \(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)

\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)

\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)

\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)

Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)

\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)

\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)

\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)

\(\Rightarrow2016S< 2017\)

\(\Rightarrow S< \frac{2017}{2016}\)

\(\Rightarrow2016A< \frac{2017}{2016}\)

\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)

ta có B= 1/2018+2/2017+3/2016+...+2017/2+2018/1

=> B=1+1+1+..+1( 2018 số hạng 1)+ 1/2018+..+2017/2

=> B= (1+1/2018)+(1+2/2017)+(1+3/2016)+...+(1+2017/2)+ 2019/2019

=> B= 2019 *(1/2+1/3+...+1/2019)

=> A/B= (1/2+1/3+...+1/2019)/2019*(1/2+1/3+..+1/2019)

=> A/B= 1/2019

Mình giúp bạn nha!

A = 2017/1 + 2017/2 + 2017/3 + . . . + 2017/2018   /   2017/1 + 2016/2 + 2015/3 + . . .+ 1/2017

    = 2017 . ( 1 + 1/2 + 1/3 + . . . +1/2018 )   /   ( 2017 . 2016 . 2015 . . . 1) . ( 1 + 1/2 + 1/3 +. . . + 1/2017 )

    = 1/2016 . 2015 . 2014. . . 1

k mình nha

Dễ mà, bạn hãy suy nghĩ đi

B = \(\frac{3^2}{2.4}+\frac{3^2}{4.6}+\frac{3^2}{6.8}+...+\frac{3^2}{198.200}\)

B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{3^2}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{3^2}{2}.\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{3^2}{2}.\left(\frac{1}{198}-\frac{1}{200}\right)\)

B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{198}-\frac{1}{200}\right)\)

B = \(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{200}\right)\)

B = \(\frac{9}{2}.\frac{99}{200}\)

B = \(\frac{891}{400}\)

D = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 48 x 49

3D = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + 4 x 5 x 3 + ... + 48 x 49 x 3

3D = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + 4 x 5 x ( 6 - 3 ) + ... + 48 x 49 x ( 50 - 47 )

3D = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 48 x 49 x 50 - 47 x 48 x 49

3D = 48 x 49 x 50

D = ( 48 x 49 x 50 ) : 3

D = 39200

E = 1² + 2² + 3² + ... + 48²

E = 1 x 1 + 2 x 2 + 3 x 3 + ... + 48 x 48

E = 1 x ( 2 - 1 ) + 2 x ( 3 - 1 ) + 3 x ( 4 - 1 ) + ... + 48 x ( 49 - 1 )

E = 1 x 2 - 1 + 2 x 3 - 2 + 3 x 4 - 3 + ... + 48 x 49 - 49

E = ( 1 x 2 + 2 x 3 + 3 x 4 + ... + 48 x 49 ) - ( 1 + 2 + 3 + ... + 49 )

Ta tính được vế trong ngoặc thứ nhất là 39200 , còn vế trong ngoặc thứ hai là 1225

thay vào ta được :

E = 39200 - 1225

E = 37975 

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Nguyễn Tiến Đạt

a)\(|3x-5|=|x+2|\)

=> Ta có 2 trường hợp

*) TH1: 3x-5=x+2

=>3x-x=2+5

=>2x=7

=>x=7:2\(\Rightarrow x=\frac{7}{2}\)

*)TH2: -3x+5=x+2

\(\Rightarrow5-3x=x+2\)

\(\Rightarrow5-2=x+3x\)

\(\Rightarrow3=4x\)

\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)

Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)

id nhu 1 tro dua

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)

Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)

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

          \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)

    \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)\)

          \(A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2016}}-\frac{1}{2^{2017}}\)

         \(A=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)

Vậy   \(A=\frac{2^{2018}-1}{2^{2017}}\)

A=đã cho.

2A=1+1/2+1/2^2+1/2^3+...+1/2^2016.

2A-A=1-1/2^2017(khử).

A=1-1/2^2017.