Tính bằng cách thuận tiện nhất:
a)13x15-150+97x15
b)2016/2015 x 4/7 -4/7 x1/2015 +3/7
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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
( 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
\(\dfrac{4}{5}\times\dfrac{3}{7}\times\dfrac{4}{5}\times\dfrac{6}{7}\times\dfrac{4}{14}=\dfrac{4}{5}\times\dfrac{3}{7}+\dfrac{4}{5}\times\dfrac{6}{7}-\dfrac{4}{5}\times\dfrac{2}{7}=\dfrac{4}{5}\times\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right)=\dfrac{4}{5}\)
\(\dfrac{3}{5}\times\dfrac{2}{7}-\dfrac{3}{5}\times\dfrac{4}{7} +\dfrac{3}{5}\)
\(=\dfrac{3}{5}\times\left(1+\dfrac{2}{7}-\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\left(\dfrac{7}{7}+\dfrac{2}{7}-\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\left(\dfrac{7+2-4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{5}{7}\)
\(=\dfrac{3}{7}\)
= 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
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