SJ
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
SJ
1
AM
14 tháng 6 2015
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
XO
13 tháng 8 2020
a) 563 x 23 + 23 x 28 - 23
= 23 x (563 + 28 - 1)
= 23 x 600
= 19200
b) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)
c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{10}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times....\times\frac{9}{10}\)
\(=\frac{1\times2\times3\times...\times9}{2\times3\times4\times...\times10}=\frac{1}{10}\)
d) \(\frac{2}{7}\times\frac{3}{9}+\frac{2}{7}\times\frac{2}{3}=\frac{2}{7}\times\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\times1=\frac{2}{7}\)
NH
13 tháng 8 2020
a ) \(563\cdot23+23\cdot28-23\)
\(=23\cdot\left(563+28-1\right)\)
\(=23\cdot600\)
\(=19200\)
b ) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)
c ) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot9}{2\cdot3\cdot4\cdot...\cdot10}=\frac{1}{10}\)
d ) \(\frac{2}{7}\cdot\frac{3}{9}+\frac{2}{7}\cdot\frac{2}{3}=\frac{2}{7}\cdot\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\cdot1=\frac{2}{7}\)
NB
0
TP
6 tháng 8 2017
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
hơi sai đề