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\(\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{2014}\right)\)
A = \(\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2012}{2013}x\frac{2013}{2014}\)
A = \(\frac{2x3x4x...x2012x2013}{3x4x5x...x2013x2014}\)
a = \(\frac{2}{2014}=\frac{1}{1007}\)
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
a, \(\frac{4}{5}-\frac{1}{3}x=\frac{1}{2}\Rightarrow\frac{1}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
\(x=\frac{3}{10}.3=\frac{9}{10}\)
Vậy, \(x=\frac{9}{10}\)
b, \(2014:x=26\left(r=12\right)\) (r là số dư nhé bạn)
Vì \(2014:x=26\) dư \(12\Rightarrow\left(2014-12\right)⋮26\)
\(\Rightarrow2002:x=26\Rightarrow x=\frac{2002}{26}=77\)
Vậy, \(x=77\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\)\(\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(15+5x=50\)
\(5x=35\)
\(x=7\)
Vậy \(x=7\)
\(\left(1+x\right)+\left(2+x\right)+\left(3+x\right)+\left(4+x\right)+\left(5+x\right)=10\times5\)
\(\Rightarrow1+x+2+x+3+x+4+x+5+x=50\)
\(\Rightarrow\left(1+2+3+4+5\right)+\left(x+x+x+x+x\right)=50\)
\(\Rightarrow15+5x=50\)
\(\Rightarrow5x=50-15\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=35:5\)
\(\Rightarrow x=7\).
0,44 x (\(x+x\times\) 5 - \(\dfrac{23}{55}\)) + \(\dfrac{3}{14}\) x 2,24 = 1
0,44 x (\(x+x\times\)5 - \(\dfrac{23}{55}\)) + 0,48 = 1
0,44 x (\(x+x\times\) 5 - \(\dfrac{23}{55}\)) = 1 - 0,48
0,44 x (\(x+x\times5\) - \(\dfrac{23}{55}\)) = 0,52
\(x+x\times5\) - \(\dfrac{23}{55}\) = 0,52 : 0,44
\(x\) x (1 + 5) -\(\dfrac{23}{55}\) = \(\dfrac{13}{11}\)
\(x\) x 6 - \(\dfrac{66}{10}\) = \(\dfrac{13}{11}\) + \(\dfrac{23}{55}\)
\(x\) x 6 = \(\dfrac{65}{55}\) + \(\dfrac{23}{55}\)
\(x\times\) 6 = \(\dfrac{8}{5}\)
\(x\) x 6 = \(\dfrac{8}{5}\) : 6
\(x\) = \(\dfrac{4}{15}\)
Vậy \(x=\dfrac{4}{15}\)