\(a,\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)

\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=13\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{7.9}\right)\)

\(=13\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(=13.\frac{2}{9}=\frac{26}{9}\)

\(b,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

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

\(=1-\frac{1}{2018}=\frac{2017}{2018}\)

P/s :Dấu chấm là dấu nhân nha

phần c đâu bn

\(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)

\(=\frac{13.10101}{15.10101}+\frac{13.10101}{35.10101}+\frac{13.10101}{63.10101}+\frac{13.10101}{99.10101}\)

\(=13.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(=\frac{13}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{13}{2}.\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=\frac{13}{2}.\frac{8}{33}\)

\(=\frac{52}{33}\)

 \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

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

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

Sửa đề chút:

 \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2018}\)

\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)

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

B tự làm nốt nhé

Ta có: 1+(1+2)+(1+2+3)+...+(1+2+3+...+2017)=2017x1+2016x2+2015x3+...+2x2016+1x2017

=> K-2016=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017x1+2016x2+2015x3+...+2x2016+1x2017}\)=\(\frac{2017x1+2016x2+2015x3+...+2x2016+1x2017}{2017x1+2016x2+2015x3+...+2x2016+1x2017}=1\)

=> K=2016+1=2017

Toán tiếng anh hả bạn

Bài này thì bạn mình có thể giải được

Thank you

At the speed of light không trả lời mà cũng được k

a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)

\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)

b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)

\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)

\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)

c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)

                 \(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)

mà \(2017.2018< 2018.2019\)

\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)

d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)

                 \(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)

mà \(2017.2018+1< 2018.2019+1\)

\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)

\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)

\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)

\(\frac{2018\times2017-1}{2016\times2018+2017}\)

\(=\frac{2018\times\left(2016+1\right)-1}{2016\times2018+2017}\)

\(=\frac{2018\times2016+2018-1}{2016\times2018+2017}\)

\(=\frac{2018\times2016+2017}{2016\times2018+2017}\)

\(=1\)

Kết quả : \(=1\)

Đáp án là 1 !

Tử số bằng mẫu số 

K-2016=1

K=2017

Muốn biết tại sao tử= mẫu thì tích nha

\(K-2016=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\times1+2016\times2+2015\times3+...+2\times2016+1\times2017}\)

\(K-2016=\frac{1\times2017+2\times2016+3\times2015+...+2017\times1}{2017\times1+2016\times2+2015\times3+...+2017\times1}\)

\(K-2016=1\)

\(\Rightarrow K=1+2016\)

\(\Rightarrow K=2017\)

bài ra 2017

 \(Ta\)có :\(a\)=\(\frac{2017\cdot2018-1}{2017.2018}\)=\(\frac{2017.2018}{2017.2018}\)-\(\frac{1}{2017.2018}\)=1-\(\frac{1}{2017.2018}\)

          \(b\)=\(\frac{2019.2020-1}{2019.2020}\)=\(\frac{2019.2020}{2019.2020}\)-\(\frac{1}{2019.2020}\)=1-\(\frac{1}{2019.2020}\)

Vì \(\frac{1}{2018.2019}\)> \(\frac{1}{2019.2020}\)nên \(a\)< \(b\)(sử dụng phần bù)

so sánh a và b biết a=2017×2018−12017×20182017×20182017×2018−1​và b =2019×2020−12019×20202019×20202019×2020−1​
 

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