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3 \(\times\) \(\dfrac{4}{15}\) + 2 \(\times\) \(\dfrac{4}{15}\) - 5 \(\times\) \(\dfrac{4}{15}\)
= \(\dfrac{4}{15}\) \(\times\) ( 3 + 2 - 5)
= \(\dfrac{4}{15}\) \(\times\) 0
= 0
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= 5 x 7 + 5 x 4 + 5 x 8 + 5 x 1
= 5 x ( 7 + 4 + 8 + 1 )
= 5 x 20
= 100
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\(\dfrac{3}{5}\times\dfrac{2}{7}-\dfrac{3}{5}\times\dfrac{4}{7} +\dfrac{3}{5}\)
\(=\dfrac{3}{5}\times\left(1+\dfrac{2}{7}-\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\left(\dfrac{7}{7}+\dfrac{2}{7}-\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\left(\dfrac{7+2-4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{5}{7}\)
\(=\dfrac{3}{7}\)
= 3/5 x 2/7 - 3/5 x 4/7 + 3/5 x 1
= 3/5 x ( 2/7 + 4/7 + 1 )
= 3/5 x 1
= 3/5
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(Dấu . là dấu nhân)
a/\(\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}:3\)
\(=\dfrac{2}{5}\cdot\dfrac{4}{3}-\dfrac{2}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\cdot1\)
\(=\dfrac{2}{5}\)
b/\(\dfrac{2010}{2018}:\dfrac{1}{2}+\dfrac{7}{2018}:\dfrac{1}{2}\)
\(=\left(\dfrac{2010}{2018}+\dfrac{7}{2018}\right):\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}:\dfrac{1}{2}\)
\(=\dfrac{2017}{2018}\cdot2\)
\(=\dfrac{2017}{1009}\)
a, \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) : 3
= \(\dfrac{2}{5}\) \(\times\) \(\dfrac{4}{3}\) - \(\dfrac{2}{5}\) \(\times\) \(\dfrac{1}{3}\)
= \(\dfrac{2}{5}\) \(\times\) ( \(\dfrac{4}{3}\) - \(\dfrac{1}{3}\))
= \(\dfrac{2}{5}\) \(\times\) 1
= \(\dfrac{2}{5}\)
b, \(\dfrac{2010}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{7}{2018}\) : \(\dfrac{1}{2}\) + \(\dfrac{1}{2018}\) : \(\dfrac{1}{2}\)
= \(\dfrac{2010}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{7}{2018}\) \(\times\) \(\dfrac{2}{1}\) + \(\dfrac{1}{2018}\) \(\times\) \(\dfrac{2}{1}\)
= \(\dfrac{2}{1}\) \(\times\) ( \(\dfrac{2010}{2018}\) + \(\dfrac{7}{2018}\) + \(\dfrac{1}{2018}\))
= 2 \(\times\) \(\dfrac{2018}{2018}\)
= 2 \(\times\) 1
= 2
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\(108\times5+108\times96-108=108\times\left(5+96-1\right)\)
\(=108\times100=10800\)
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\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
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\(2005-5+5+5+...+5+5\)
\(=2005-5+\left(5+5+...+5\right)\)
\(=2005-5+50\)
\(=2000+50=2050\)
PP/ss: Chắc z ((:
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= 5 x 107 + 5 x 1 - 5 x 8/ 5 x 1
= 5 x ( 107+1-8)/ 5 x 1
= 5 x 100/5 x 1
= 1 x 100/ 5 x 1
= 100
3x4/5+2x4/5-5x4/5
=4/5x[3+2-5]
4/5x0
=0