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Gọi tổng dãy số hạng trên là A
A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)
Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số
A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\)
A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)
= \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{90}\right)\)
= \(\left(1+1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)(9 số 1)
= 9 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
= \(9+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
= \(9+\left(1-\frac{1}{10}\right)=9+\frac{9}{10}=\frac{90}{10}+\frac{9}{10}=\frac{99}{10}\)
\(y=\frac{4\frac{6}{11}x11\frac{8}{9}+4\frac{12}{13}:3\frac{2}{5}}{123\frac{34}{45}:21\frac{1}{8}}\)
a ) \(3\frac{4}{5}-2\frac{3}{4}:1\frac{1}{8}=\frac{19}{5}-\frac{11}{4}:\frac{9}{8}=\frac{19}{5}-\frac{22}{9}=\frac{61}{45}\)
b ) \(4\frac{5}{7}:1\frac{5}{6}+2\frac{7}{15}.\frac{21}{74}=\frac{33}{7}:\frac{11}{6}+\frac{37}{15}.\frac{21}{74}=\frac{18}{7}+\frac{7}{10}=\frac{229}{70}\)
Ta có:
\(\frac{2}{3}+\frac{4}{21}+\frac{6}{91}+\frac{8}{273}+\frac{3}{504}\)
\(=\frac{2}{1.3}+\frac{4}{3.7}+\frac{6}{7.13}+\frac{8}{13.21}+\frac{3}{21.24}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{21}-\frac{1}{24}\)
\(=1-\frac{1}{24}=\frac{23}{24}\)
Vậy giá trị biểu thức là \(\frac{23}{24}\)