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1)
A = \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}\)
= \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}\)
= \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}\)
= \(\frac{1}{5}-\frac{1}{12}\)
= \(\frac{7}{60}\)
B = \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
= \(\frac{3.4.5.....100}{2.3.4....99}\)
= \(\frac{100}{2}=50\)
C = \(\frac{1}{4^{2-1}}+\frac{1}{6^{2-1}}+\frac{1}{8^{2-1}}...+\frac{1}{30^{2-1}}\)
= \(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{30}\)
= \(\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{2.15}\)
= \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{2}.\frac{1}{4}+...+\frac{1}{2}.\frac{1}{15}\)
= \(\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{15}\right)\)
\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+\left(\frac{1}{10}-\frac{1}{10}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
~ Hok tốt ~
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\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}-\frac{1}{256}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}\)
\(A+2A=\left(\frac{1}{2}-\frac{1}{4}+...-\frac{1}{256}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-...-\frac{1}{128}\right)\)
\(3A=1-\frac{1}{256}< 1\)
\(\Rightarrow A< \frac{1}{3}\).
a, \(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{999}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{1000}{999}\)
\(=\dfrac{3.4.5...1000}{2.3.4...999}\)
\(=\dfrac{1000}{2}\)\(=500\)
b, \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1000}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-999}{1000}\)
\(=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-999\right)}{2.3.4...1000}\)
\(=\dfrac{-1}{1000}\)
