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

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

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

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

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

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

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

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

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

\(B=\frac{1}{1.2}=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

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

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

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

Vậy \(B=\frac{2017}{2018}\)

rút gọn là được mà ! 

\(\frac{1}{2}\)

\(\frac{7}{14}\)

\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+...+\dfrac{2}{48\times49\times50}\)

\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+...+\dfrac{1}{48\times49}-\dfrac{1}{49\times50}\)

\(=\dfrac{1}{1\times2}-\dfrac{1}{49\times50}\)

\(=\dfrac{1}{2}-\dfrac{1}{2450}\)

\(=\dfrac{612}{1225}\)

\(\text{#}Toru\)

1/2 - 1/49 x50

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\) 

Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
 =1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050

Giúp mình với tối nay mình học rồi!!!

Mik làm gộp lại nghe bạn!!

Ta có : S = \(\frac{5.2^{30}.6^3.3^{15}-2^3.8^9.3^{17}.21}{21.2^{29}.3^{16}.4-2^{29}.\left(3^4\right)^5}=\frac{5.2^{30}.\left(2.3\right)^3.3^{15}-2^3.\left(2^3\right)^9.3^{17}.3.7}{3.7.2^{29}.3^{16}.2^2-2^{29}.3^{20}}=\frac{5.2^{33}.3^{18}-2^{30}.3^{18}.7}{3^{17}.7.2^{31}-2^{29}.3^{20}}\)

\(=\frac{2^{30}.3^{18}.\left(5.2^3-7\right)}{3^{17}.2^{29}.\left(7.2^2-3^3\right)}=2.3.33=198\)

\(=\frac{1.2}{99.100}\)

\(=\frac{2}{9900}=\frac{1}{4950}\)