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
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)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
NT
0
H9
HT.Phong (9A5)
CTVHS
2 tháng 11 2023
a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{7}\times1\)
\(=\dfrac{5}{7}\)
b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)
\(=1+0\)
\(=1\)
c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)
\(=\dfrac{5}{7}+\dfrac{2}{7}\)
\(=1\)
d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)
\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)
\(=\dfrac{1}{2}+1\)
\(=\dfrac{3}{2}\)
e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)
\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{12}{10}\)
\(=\dfrac{4}{5}+\dfrac{6}{5}\)
\(=\dfrac{10}{5}\)
\(=2\)
g) \(362\times728+326\times272\)
\(=326\times\left(728+272\right)\)
\(=326\times1000\)
\(=326000\)
T
19 tháng 9 2023
\(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+\cdot\cdot\cdot+\dfrac{2}{97\times100}\)
\(=\dfrac{2}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\cdot\cdot\cdot+\dfrac{3}{97\times100}\right)\)
\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\cdot\cdot\cdot+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\times\dfrac{99}{100}\)
\(=\dfrac{33}{50}\)
#\(Toru\)
VH
1 tháng 4 2023
Bài 2. > , < , = ?
5/7 < 4/3 2/5 < 6/10 1/4 = 3/12 27/36 >
2/9 7/6 > 7/9
7/2 = 2/7 15/23 < 1 27/9 > 2 14/15 < 1 51/17 < 4
TT
0
Sửa đề:
\(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)
\(=2\times\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)
\(=\dfrac{2}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+...+\dfrac{3}{97\times100}\right)\)
\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\times\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{2}{3}\times\dfrac{99}{100}\)
\(=\dfrac{198}{300}\)
\(=\dfrac{33}{50}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)