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\(A=\frac{2}{2.4.6}+\frac{2}{4.6.8}+\frac{2}{6.8.10}+\frac{2}{8.10.12}\)
\(A=\frac{2}{48}+\frac{2}{192}+\frac{2}{480}+\frac{2}{960}\)
\(A=\frac{1}{24}+\frac{1}{96}+\frac{1}{240}+\frac{1}{480}\)
\(A=\frac{20}{480}+\frac{5}{480}+\frac{2}{480}+\frac{1}{480}\)
\(A=\frac{7}{120}\)
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 = \(\dfrac{7}{120}\)
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Ta nhận thấy
\(\dfrac{1}{n\cdot\left(n+2\right)}-\dfrac{1}{\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{n+4}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}-\dfrac{n}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{n+4-n}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{4}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\)
\(A=\dfrac{4}{2\cdot4\cdot6}+\dfrac{4}{4\cdot6\cdot8}+\dfrac{4}{6\cdot8\cdot10}+...+\dfrac{4}{46\cdot48\cdot50}\\ =\dfrac{1}{2\cdot4}-\dfrac{1}{4\cdot6}+\dfrac{1}{4\cdot6}-\dfrac{1}{6\cdot8}+\dfrac{1}{6\cdot8}-\dfrac{1}{8\cdot10}+...+\dfrac{1}{46\cdot48}-\dfrac{1}{48\cdot50}\\ =\dfrac{1}{2\cdot4}-\dfrac{1}{48\cdot50}\\ =\dfrac{1}{8}-\dfrac{1}{2400}\\ =\dfrac{300}{2400}-\dfrac{1}{2400}\\ =\dfrac{299}{2400}\)
Số nghịch đảo của \(A\) là \(\dfrac{2400}{299}\)
A = \(\frac{2.4.6+4.6.8+6.8.10+8.10.12+...+198.200.202}{1.3.5+3.5.7+5.7.9+7.9.11+...+97.99.101}\) =?
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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 = \(\frac{1}{2}\)
Tất nhiên là < hơn 2 rùi nhưng mk quên ùi
2 lớn hơn phải không