1)1/4+1/12+1/24+1/40+1/60+1/84
2)tìm x
1/2011x=(1-1/2)(1-1/3)(1-1/4)......(1-1/2010)(1-1/2011)
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Ta có \(B=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{2}{2010}+1\right)+\left(\frac{1}{2011}+1\right)+1\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2010}+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)\)
B=2012.A
=>A/B=1/2012
Sửa đề:
\(\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\dfrac{4-3-1}{12}\)
\(=\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\dfrac{0}{12}\)
\(=0\)
2A=1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
\(\dfrac{1}{4}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{40}\) + \(\dfrac{1}{60}\) + \(\dfrac{1}{84}\) + \(x\) = 1
\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.6}\)+\(\dfrac{1}{2.12}\)+\(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.30}\) + \(\dfrac{1}{2.42}\) + \(x\) =1
\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\) + \(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).\(\dfrac{6}{7}\) + \(x\) = 1
\(\dfrac{3}{7}\) + \(x\) = 1
\(x\) = 1 - \(\dfrac{3}{7}\)
\(x\) = \(\dfrac{4}{7}\)
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84} \)
\(=\frac{210}{840}+\frac{70}{840}+\frac{35}{840}+\frac{21}{840}+\frac{14}{840}+\frac{10}{840}\)
\(=\frac{210+70+35+21+14+10}{840}\)
\(=\frac{360}{840}\)
\(=\frac{3}{7}\)