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Ta có:
\(3^3=27\text{≡}1\left(mod13\right)\)
\(\Rightarrow\left(3^3\right)^{33}\text{≡}1^{33}\left(mod13\right)\)
\(\Rightarrow3^{99}\text{≡}1\left(mod13\right)\)
\(\Rightarrow3^{100}\text{≡}3\left(mod13\right)\)
Lại có :
\(\left(3^3\right)^{35}\text{≡}1^{35}\left(mod13\right)\)
\(\Rightarrow3^{105}\text{≡}1\left(mod13\right)\)
\(\Rightarrow3^{100}+3^{105}\text{≡}3+1=4\left(mod13\right)\)
Vậy số dư trong phép chia đó là 4.
nao giai ho minh
\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}\) + \(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}\)
= \(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3.\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}\) + \(\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}\)
= \(\frac{1}{3}\) + \(\frac{2}{3}\) = 1
H=\(\frac{5}{28}\) + \(\frac{5}{70}\) + \(\frac{5}{130}\) +...+ \(\frac{5}{700}\)
H= \(\frac{5}{4.7}\) + \(\frac{5}{7.10}\) + \(\frac{5}{10.13}\) +...+\(\frac{5}{25.28}\)
H= \(\frac{5}{3}\) (\(\frac{1}{4}\) - \(\frac{1}{7}\) + \(\frac{1}{7}\) - \(\frac{1}{10}\) + \(\frac{1}{10}\) - \(\frac{1}{13}\) +...+ \(\frac{1}{25}\) - \(\frac{1}{28}\))
H= \(\frac{5}{3}\) (\(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) - \(\frac{1}{10}\)+...+ \(\frac{1}{25}\)- \(\frac{1}{25}\) - \(\frac{1}{28}\))
H= \(\frac{5}{3}\) ( \(\frac{1}{4}\) - \(\frac{1}{28}\)) = \(\frac{5}{3}\) . \(\frac{3}{14}\)= \(\frac{5}{14}\)
