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

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

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

\(=\frac{2017}{2018}\)  

help me

\(a)\) Ta có : 

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

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)

\(\Leftrightarrow\)\(x+1=2019\)

\(\Leftrightarrow\)\(x=2019-1\)

\(\Leftrightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~ 

id nhu 1 tro dua

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

Trịnh Lê Na làm kiểu j mà chẳng hiểu thế

\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)

\(A=2018\)

Ta có : 

\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)

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

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

\(A=2018\)

Vậy \(A=2018\)

Chúc bạn học tốt ~ 

mình giỏi vl vậy >:

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 

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

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

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

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