1 nhân 1=vỏiehough9[ruayt8a[u9tui4yah5uhg9i5[hgioht[i95hyiruahuirpjyruhpayierhyei
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\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
\(x=\dfrac{9}{12}-\dfrac{2}{12}\)
\(x=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)
\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)
\(x=\dfrac{1009}{2020}\)
\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)
\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)
\(=\dfrac{2}{5}\times\dfrac{3}{7}\)
\(=\dfrac{6}{35}\)
\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)
= 1/2 x 2/3 x 3/4 x ..... x 2014/2015
= 1 x 2 x 3 x ... x 2014/2 x 3 x 4 x .... x 2015
= 1/2015
k mk nha
A = (1 - \(\frac{1}{2}\)) x (1 - \(\frac{1}{3}\)) x (1 - \(\frac{1}{4}\)) x (1 - \(\frac{1}{5}\)) x ... x (1 - \(\frac{1}{2014}\)) x (1 - \(\frac{1}{2015}\))
A = \(\frac{1}{2}\)x \(\frac{2}{3}\) x \(\frac{3}{4}\) x \(\frac{4}{5}\) x ... x \(\frac{2013}{2014}\)x \(\frac{2014}{2015}\)
A = \(\frac{1x2x3x4x...x2013x2014}{2x3x4x5x...x2014x2015}\)
A = \(\frac{1}{2015}\)
Vậy A = \(\frac{1}{2015}\)
~~~
\(a,\left(1-\dfrac{1}{10}\right)\times\left(1-\dfrac{1}{11}\right)\times\left(1-\dfrac{1}{12}\right)\times\left(1-\dfrac{1}{13}\right)\times\left(1-\dfrac{1}{14}\right)\times\left(1-\dfrac{1}{15}\right)=\dfrac{9}{10}\times\dfrac{10}{11}\times\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{15}=\dfrac{9}{15}=\dfrac{3}{5}\)
\(b,\dfrac{2013\times2012-2}{2011+2011\times2013}=\dfrac{\left(2014-1\right)\times2012-2}{2011\times\left(2013+1\right)}=\dfrac{2014\times2012-2012-2}{2011\times2014}=\dfrac{2014\times2012-2014}{2011\times2014}=\dfrac{2014\times\left(2012-1\right)}{2011\times2014}=\dfrac{2011\times2014}{2011\times2014}=1\)
tính bằng cách thuân tiên nhất
(1-1/2) nhân (1-1/3) nhân (1-1/4) nhân (1-1/5) nhân (1-1/6)
a) (456x35+65x456):19
=456x(65+35):19
=456x100:19
=45600:19
=2400
k mình nhé
Chúc bạn học giỏi
\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016\cdot2017}{2\cdot3\cdot4\cdot...\cdot2017\cdot2018}\)
\(C=\frac{1}{2018}\)
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cho nó vô trong này nè