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a, 13 \(\times\) 15 - 150 + 97 \(\times\)15
13 \(\times\) 15 - 15 \(\times\) 10 + 97 \(\times\) 15
= 15 \(\times\) ( 13 - 10 + 97)
= 15 \(\times\) ( 3 + 97)
= 15 \(\times\) 100
=1500
b, \(\dfrac{2016}{2015}\) \(\times\) \(\dfrac{4}{7}\) - \(\dfrac{4}{7}\) \(\times\) \(\dfrac{1}{2015}\) + \(\dfrac{3}{7}\)
= \(\dfrac{4}{7}\) \(\times\) ( \(\dfrac{2016}{2015}\) - \(\dfrac{1}{2015}\)) + \(\dfrac{3}{7}\)
= \(\dfrac{4}{7}\) \(\times\) \(\dfrac{2015}{2015}\) + \(\dfrac{3}{7}\)
= \(\dfrac{4}{7}\) + \(\dfrac{3}{7}\)
= \(\dfrac{7}{7}\)
= 1
( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 1 + 1/3 - 4/3)
=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 4/3 - 4/3)
=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x 0
=0
Ta có: \(\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(=\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(\dfrac{3}{3}+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
=0
\(\dfrac{7}{19}\times\dfrac{1}{3}+\dfrac{7}{13}\times\dfrac{2}{3}\)
\(=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{19}\times1\)
\(=\dfrac{7}{19}\)
( 2014 x 2015 - 2016 ) : ( 2012 + 2013 x 2014 )
= ( 4058210 - 2016 ) : ( 2012 + 4054182 )
= 4056194 : 4056194
= 1
a, \(13\) \(\times\) 15 - 150 + 97 \(\times\) 15
= 15 \(\times\) ( 13 - 10 + 97)
= 15 \(\times\) ( 3 + 97)
= 15 \(\times\) 100
= 1500
b, \(\dfrac{2016}{2015}\) \(\times\) \(\dfrac{4}{7}\) - \(\dfrac{4}{7}\) \(\times\) \(\dfrac{1}{2015}\) + \(\dfrac{3}{7}\)
= \(\dfrac{4}{7}\) \(\times\) ( \(\dfrac{2016}{2015}\) - \(\dfrac{1}{2015}\)) + \(\dfrac{3}{7}\)
= \(\dfrac{4}{7}\) \(\times\) \(\dfrac{2015}{2015}\) + \(\dfrac{3}{7}\)
= \(\dfrac{4}{7}\) + \(\dfrac{3}{7}\)
= \(\dfrac{7}{7}\)
= 1