Adfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1
dfrac{2n+5}{n+2}
Sdfrac{1}{20}+dfrac{1}{21}+dfrac{1}{22}+L+dfrac{1}{29}
Adfrac{7}{4}.left(dfrac{3333}{1212}+dfrac{3333}{2020}+dfrac{3333}{3030}+dfrac{3333}{4242}right)
Adfrac{1}{4.5}+dfrac{1}{5.6}+...+dfrac{1}{48.49}+dfrac{1}{49.50}
Adfrac{17}{1.3}+dfrac{17}{3.5}+dfrac{17}{5.7}+...+dfrac{17}{49.51}
dfrac{2n+5}{3n+1}
left(left|xright|+dfrac{2}{5}right):dfrac{2}{5}1
dfrac{1}{2}+dfrac{1}{4}+dfrac{1}{8}+dfrac{1}{16}+dfra...
Đọc tiếp
\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(\dfrac{2n+5}{n+2}\)
\(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+L+\dfrac{1}{29}\)
\(A=\dfrac{7}{4}.\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}\)
\(A=\dfrac{17}{1.3}+\dfrac{17}{3.5}+\dfrac{17}{5.7}+...+\dfrac{17}{49.51}\)
\(\dfrac{2n+5}{3n+1}\)
\(\left(\left|x\right|+\dfrac{2}{5}\right):\dfrac{2}{5}=1\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}< 1\)
\(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
\(5.\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{10}\right)\le\dfrac{x}{20}\le-3\left(\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{5}\right)\left(x\in Z\right)\)
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}< 1\)
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