(\(\frac{1}{21}\)+\(\frac{1}{210}\)+\(\frac{1}{2010}\))*(\(\frac{1}{3}\)-\(\frac{1}{30}\)-\(\frac{1}{5}\)-\(\frac{1}{10}\))
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KT
19 tháng 10 2017
Ta thấy: \(\frac{1}{3}\)- \(\frac{1}{30}\)- \(\frac{1}{5}\)- \(\frac{1}{10}\)
= \(\frac{10}{30}\)- \(\frac{1}{30}\)- \(\frac{6}{30}\)- \(\frac{3}{30}\)= 0
nên gtbt trên bằng 0
NT
4
4 tháng 3 2017
Ta có:
\(\frac{x}{2013}\)-\(\frac{1}{10}\)-\(\frac{1}{15}\)-\(\frac{1}{21}\)-...-\(\frac{1}{120}\)=\(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- (\(\frac{2}{20}\)+\(\frac{2}{30}\)+\(\frac{2}{42}\)+...+\(\frac{2}{240}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{15.16}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4}\)-\(\frac{1}{10}\)) = \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)- 2.\(\frac{3}{10}\)= \(\frac{5}{8}\)
=>\(\frac{x}{2013}\)= \(\frac{5}{8}\)+\(\frac{6}{10}\)= 1
=> \(x=2013\)
Vậy \(x=2013\)
TV
11 tháng 2 2020
a)\(\frac{5}{21}\)+\(\frac{-3}{7}\)<\(\frac{x}{21}\)<\(\frac{-2}{7}\)+\(\frac{8}{21}\)
\(\Rightarrow\)\(\frac{-4}{21}\)<\(\frac{x}{21}\)<\(\frac{2}{21}\)
\(\Rightarrow\)\(\frac{x}{21}\)\(\in\)\(\left\{\frac{-3}{21};\frac{-2}{21};\frac{-1}{21};\frac{0}{21};\frac{1}{21}\right\}\)
vậy x\(\in\)\(\left\{-3;-2;-1;0;1\right\}\)
28 tháng 5 2018
\(\Rightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{6}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{5}{21}-\frac{1}{3}\)
\(\Rightarrow\frac{1}{42}\cdot\frac{x}{3}=\frac{-2}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{-2}{21}\div\frac{1}{42}\)
\(\Rightarrow\frac{x}{3}=-4\)
\(\Rightarrow\frac{x}{3}=\frac{-12}{3}\)
\(\Rightarrow x=-12\)
30 tháng 1 2022
a: \(=\dfrac{1}{280}-\dfrac{1}{70}=\dfrac{1}{280}-\dfrac{4}{280}=-\dfrac{3}{280}\)
b: \(=\dfrac{200}{37}\left(\dfrac{85}{26}-\dfrac{59}{26}\right)=\dfrac{200}{37}\)
c: \(=2+\dfrac{2}{2+\dfrac{2}{2+\dfrac{2}{\dfrac{5}{2}}}}=2+\dfrac{2}{2+\dfrac{2}{2+\dfrac{4}{5}}}=2+\dfrac{2}{2+\dfrac{5}{7}}\)
\(=2+2:\dfrac{19}{7}=2+\dfrac{14}{19}=\dfrac{38+14}{19}=\dfrac{52}{19}\)
Ta có:
\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10}{30}-\frac{1}{30}-\frac{6}{30}-\frac{3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10-1-6-3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(0\)
= \(0\)
ta có \(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}=\frac{10-1-6-3}{30}=\frac{0}{30}=0\)
=>\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)x\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)=0\)