Tách 2019 thành 2018+1

     \(2018\times2018-2019\times2017\)

\(=2018\times\left(2017+1\right)-\left(2018+1\right)\times2017\)

\(=2018\times2017+2018-2018\times2017-2017\)

\(=2018\times2017-2018\times2017+2018-2017\)

\(=2018-2017\)

\(=1\)    

                       ~~~~~~~~~~~Hok tốt~~~~~~~~~~~

= 1 nha bn

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

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

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

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

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

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

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

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

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

$A=\dfrac{2018.2017-1}{2016.2018+2017}$

$=>A={2018.2016+2018-1}{2016.2018+2017}$

$=>A={2018.2016+2017}{2016.2018+2017}$

$=>A=1$

\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)

\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)

\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)

\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)

\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)

\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)

\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)

Chúc bạn học tốt!!!

222

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 \(2017.2018+1< 2018.2019+1\)

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

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

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

\(\frac{2016\times2017+4034}{2018\times2019-4034}=\frac{2016\times2017+2\times2017}{2018\times2019-2\times2017}\)

\(=\frac{\left(2016+2\right)\times2017}{2018\times2017+2\times2017-2\times2017}=\frac{\left(2016+2\right)\times2017}{2018\times2017+0}\)

\(=\frac{2018\times2017}{2018\times2017}=1\)

bảng 1 nha bạn

= \(\frac{2017x2018+2}{2018x\left(2017+1\right)-2016}\)= \(\frac{2017x2018+2}{2017x2018+2018-2016}\) = \(\frac{2017x2018+2}{2017x2018+2}\) = 1

\(\frac{2016\times2018+2}{2016\times2017+2018}=\frac{2016\times\left(2017+1\right)+2}{2016\times2017+2018}=\)\(=\frac{2016\times2017+2016+2}{2016\times2017+2018}=\frac{2016\times2017+2018}{2016\times2017+2018}=1\)

Ta có : \(\frac{2016\times2018+2}{2016\times2017+2018}=\frac{2016\left(2017+1\right)+2}{2016\times2017+2018}=\frac{2016\times2017+2016+2}{2016\times2017+2018}\)

\(=\frac{2016\times2017+2018}{2016\times2017+2018}=1\)

2016x(2016+2)+2016-2014  2016x2016+2x2016+2016-2014       2019 x 2016 - 2014    2019 x 2016 - 2016 + 2   2018 x 2016+2

_____________________ =____________________________= ________________=___________________=____________= 1

2016x(2016+1)+2016+2       2016 x 2016 + 1 x 2016 + 2016 + 2         2018 x 2016 + 2         2018 x 2016 +2         2018 x 2016 +2