Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải:
a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\frac{1}{18}\)
C = \(2.\frac{1}{18}\)
C = \(\frac{1}{9}\)
Vậy C = \(\frac{1}{9}\)
b) D = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
D = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)\
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\frac{2}{75}\)
D = \(\frac{1}{75}\)
Vậy D = \(\frac{1}{75}\)
c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)
E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)
E = \(\frac{1}{8}-\frac{1}{41}\)
E = \(\frac{33}{328}\)
Vậy E = \(\frac{33}{328}\)
A=.....
=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY
TỰ LÀM NHE
\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)
\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)
\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(C=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=9-\left(1-\frac{1}{10}\right)\)
\(C=9-\frac{9}{10}=\frac{81}{10}\)
a, \(A=\frac{6}{10.11}+\frac{6}{11.12}+\frac{6}{12.13}+...+\frac{6}{69.70}\)
\(A=\frac{6}{10}-\frac{6}{11}+\frac{6}{11}-\frac{6}{12}+\frac{6}{12}-\frac{6}{13}+...+\frac{6}{69}-\frac{6}{70}\)
\(A=\frac{6}{10}-\frac{6}{70}\)
\(A=\frac{18}{35}\)
b, \(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
\(B=\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(B=2.\frac{1009}{2020}\)
\(B=\frac{1009}{1010}\)
Chúc bạn học tốt 
Hơi thắc mắc câu B cậu oi!!!Gỉai thích cho mk vs ạ!!Thanks
Mình nói lí thuyết cho nghe:
Với phân số \(\frac{a-b}{a.b}\)\(\left(VD:\frac{1}{1.2};\frac{1}{2.3};\frac{1}{2015.2016};\frac{3}{15.18};\frac{3}{18.21};\frac{1}{10.11};\frac{1}{11.12};...\right)\)thì:
\(\frac{b-a}{a.b}=\frac{b}{a.b}-\frac{a}{a.b}=\frac{1}{a}-\frac{1}{b}\left(VD:\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{3}{15.18}=\frac{1}{15}-\frac{1}{18}\right)\)
ÁP dụng để tính:
c) \(\Rightarrow\frac{1}{4}C=\frac{1}{4}\left(\frac{12}{15.18}+\frac{12}{18.21}+...+\frac{12}{87.90}\right)=\frac{3}{15.18}+\frac{3}{18.21}+....+\frac{3}{87.90}\)
\(\Rightarrow\frac{1}{4}C=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}=\frac{1}{15}-\frac{1}{90}\)
=> \(C=\left(\frac{1}{15}-\frac{1}{90}\right).4\)
a,\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=1-\frac{1}{2016}\)suy ra \(A=\frac{2015}{2016}\)
b, \(B=5\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)
\(B=5\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(B=5\left(\frac{1}{10}-\frac{1}{70}\right)\)suy ra \(B=5.\frac{3}{35}\)
\(B=\frac{3}{7}\)
c,\(C=4.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(C=4.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(C=4.\left(\frac{1}{15}-\frac{1}{90}\right)\)suy ra \(C=4.\frac{1}{18}\)
\(C=\frac{2}{9}\)
b)
\(5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(5\left(1-\frac{1}{6}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(5\left(1-\frac{1}{31}\right)\)
\(=\frac{150}{31}\)
a)
\(\frac{7}{10}-\frac{7}{11}+...+\frac{7}{69}-\frac{7}{70}\)
\(=\frac{7}{10}-\frac{7}{70}\)
\(=\frac{3}{5}\)
\(A=\frac{21}{31}+\frac{-16}{7}+\frac{44}{53}+\frac{10}{21}+\frac{9}{53} \)
\(A=\left(\frac{16}{7}+\frac{10}{21}\right)+\left(\frac{44}{53}+\frac{9}{53}\right)+\frac{21}{31}\)
\(A=\frac{58}{21}+1+\frac{21}{31}\)
\(A=\frac{100}{21}\)
\(B=6\left(\frac{1}{15.18}+\frac{1}{18.21}+...+\frac{1}{87.90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=6\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=6.\frac{1}{18}\)
\(B=\frac{1}{3}\)
2) = -1/2 . -2/3 .-3/4 ..... . -98/99 = 1/99 (Tích này có 98 thừa số âm, 98 là số chẵn nên tích mang dấu dương)