Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
mình nghĩ bạn chép ngược đề và đề đúng phải là abc1-1abc=1107, mình giải theo đề này nha:
abc.10+1-1000-abc=1107
abc.9+1-1000=1107
abc.9-999=1107
abc.9=2106
abc=2106:9
abc=234
chúc bạn học tốt nha
\(1-\left(11\frac{1}{2}-10,1+x\right):8\frac{2}{5}=0\)
=>\(1-\left(\frac{21}{2}-\frac{101}{10}+x\right):\frac{42}{5}=0\)
=> \(1-\left(\frac{2}{5}+x\right)=0\)
=>\(1=\frac{2}{5}+x\)
=>\(x=1-\frac{2}{5}\)
=>\(x=\frac{3}{5}\)
Vậy ..................
Ta có: \(M=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Leftrightarrow\frac{1}{2}M=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{6}-\frac{1}{11}=\frac{5}{66}\)
\(\Rightarrow M=\frac{5}{66}:\frac{1}{2}=\frac{5}{33}.\)
\(M=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}\)
\(M=\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}\)
\(M=\frac{2}{6\cdot7}+\frac{2}{7\cdot8}+\frac{2}{8\cdot9}+\frac{2}{9\cdot10}+\frac{2}{10\cdot11}\)
\(M=2\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)
\(M=2\left(\frac{1}{6}-\frac{1}{11}\right)\)
\(M=2\cdot\frac{5}{66}\)
\(M=\frac{5}{33}\)
\(=\frac{2017\times\left(2018-1\right)}{2018\times2016+2018-2017}\)
\(=\frac{2017\times2017}{2018\times\left(2016+1\right)-2017}\)
\(=\frac{2017\times2017}{2018\times2017-2017}\)
\(=\frac{2017\times2017}{2017\times\left(2018-1\right)}=\frac{2017\times2017}{2017\times2017}=1\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.
\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
5dam 47 dm = 50m + 4,7m = 54,7m
54,7 m