Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+\frac{7}{20}+\frac{7}{30}+\frac{7}{42}+\frac{7}{56}+\frac{7}{72}+\frac{7}{90}\)\(\frac{7}{90}\)
=\(\frac{7}{2+6+12+20+30+42+56+72+90}\)
=\(\frac{63}{10}\)
=6.3
= \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{90}\right)\)
= \(\left(1+1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)(9 số 1)
= 9 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
= \(9+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
= \(9+\left(1-\frac{1}{10}\right)=9+\frac{9}{10}=\frac{90}{10}+\frac{9}{10}=\frac{99}{10}\)
\(\frac{18}{11}+\left(\frac{7}{4}-\frac{3}{5}\right):\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}:\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}\times\frac{2}{1}\)
\(\frac{18}{11}+\frac{23}{10}\)
\(\frac{433}{110}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+\frac{3}{5.6}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\)
Gọi \(\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+......+\frac{3}{9.10}\right)\)là \(A\); \(\left(\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}\right)\)là B . Ta có :
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}.\left(\frac{1}{1}-\frac{1}{10}\right)\)
\(A=\frac{3}{1}\cdot\frac{9}{10}=\frac{27}{10}\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{6}-\frac{1}{16}+\frac{1}{16}-\frac{1}{23}+....+\frac{1}{93}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(B=\frac{77}{7}\cdot\frac{49}{100}=\frac{539}{100}\)
\(\Rightarrow\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+\frac{3}{4.5}+...+\frac{3}{9.10}+\frac{77}{2.9}+\frac{77}{9.16}+\frac{77}{16.23}+...+\frac{77}{93.100}=\frac{27}{10}+\frac{539}{100}=\frac{809}{100}\)
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}\cdot\frac{4}{15}=\frac{16}{45}\)
a)\(\frac{77}{24}\)=\(3\frac{5}{24}\)
b)=10
c)\(\frac{24}{35}\)
\(\frac{7}{2}+\frac{7}{6}+\frac{7}{12}+\frac{7}{20}+\frac{7}{30}+\frac{7}{42}=7\times\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=7\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}\right)\)
\(=7\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=7\times\left(\frac{1}{1}-\frac{1}{7}\right)=7\times\left(\frac{7}{7}-\frac{1}{7}\right)=7\times\frac{6}{7}=6\)
7\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
7\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
7\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
7\(\left(1-\frac{1}{7}\right)\)
7\(.\frac{6}{7}\)
6