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\(\dfrac{3}{2.6}\) + \(\dfrac{3}{6.10}\) + \(\dfrac{3}{10.14}\)
= \(\dfrac{3}{4}\).(\(\dfrac{4}{2.6}\) + \(\dfrac{4}{6.10}\) + \(\dfrac{4}{10.14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}-\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{14}\))
= \(\dfrac{3}{4}\). \(\dfrac{3}{7}\)
= \(\dfrac{9}{28}\)
B = \(\dfrac{4}{1.3.5}\) + \(\dfrac{4}{3.5.7}\) + \(\dfrac{4}{5.7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{3.5}\) + \(\dfrac{1}{3.5}\) - \(\dfrac{1}{5.7}\) + \(\dfrac{1}{5.7}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{1.3}\) - \(\dfrac{1}{7.9}\)
B = \(\dfrac{1}{3}\) - \(\dfrac{1}{63}\)
B = \(\dfrac{20}{63}\)
\(A=\frac{2}{2.4.6}+\frac{2}{4.6.8}+\frac{2}{6.8.10}+\frac{2}{8.10.12}\)
\(A=\frac{2}{48}+\frac{2}{192}+\frac{2}{480}+\frac{2}{960}\)
\(A=\frac{1}{24}+\frac{1}{96}+\frac{1}{240}+\frac{1}{480}\)
\(A=\frac{20}{480}+\frac{5}{480}+\frac{2}{480}+\frac{1}{480}\)
\(A=\frac{7}{120}\)
A = \(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\)
A = \(\dfrac{2}{2}\).(\(\dfrac{2}{2.4.6}\) + \(\dfrac{2}{4.6.8}\) + \(\dfrac{2}{6.8.10}\) + \(\dfrac{2}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{2.2}{2.4.6}\) + \(\dfrac{2.2}{4.6.8}\) + \(\dfrac{2.2}{6.8.10}+\dfrac{2.2}{8.10.12}\))
A = \(\dfrac{1}{2}\).( \(\dfrac{4}{2.4.6}+\dfrac{4}{4.6.8}+\dfrac{4}{6.8.10}+\dfrac{4}{8.10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{4.6}\) +\(\dfrac{1}{4.6}\) - \(\dfrac{1}{6.8}\) + \(\dfrac{1}{6.8}\) - \(\dfrac{1}{8.10}\) + \(\dfrac{1}{8.10}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{2.4}\) - \(\dfrac{1}{10.12}\))
A = \(\dfrac{1}{2}\).(\(\dfrac{1}{8}-\dfrac{1}{120}\))
A = \(\dfrac{1}{2}\).\(\dfrac{7}{60}\)
A = \(\dfrac{7}{120}\)
Ta có A = \(\frac{1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8}{2.4.6-4.6.8+6.8.10-8.10.12+10.12.14-12.14.16}\)
A = \(\frac{1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8}{\left(1.2.3\right).2-\left(2.3.4\right).2+\left(3.4.5\right).2-\left(4.5.6\right).2+\left(5.6.7\right).2-\left(6.7.8\right).2}\)
A = \(\frac{1.\left(1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8\right)}{2.\left(1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8\right)}\)
A = \(\frac{1}{2}\)
\(A=\dfrac{2.6.10+6.10.14+10.14.18+..+194.198.202}{1.3.5+3.5.7+5.7.9+..+97.99.101}\)
\(=\dfrac{2^3.1.3.5+2^3.3.5.7+2^3.5.7.9+...+2^3.97.99.101}{1.3.5+3.5.7+7.9.11+...+97.99.101}\)
\(=\dfrac{2^3.\left(1.3.5+3.5.7+7.9.11+...+97.99.101\right)}{1.3.5+3.5.7+5.7.9+...+97.99.101}=2^3=8\)
ta có tử =2.(1.3.5)+2.(3.5.7)+2.(5.7.9)+....+2.(97.99.101)
=2.(1.3.5+3.5.7+5.7.9+....+97.99.101)
mà mẫu =1.3.5+3.5.7+5.7.9+...+97.99.101
suy ra A=2