tính bằng cách thuận tiện nhất 1002 × 2015 -2015 -2015
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2015 x 8 + 9 x 2015 - 2015 x 6 - 2015
= 2015 x 8 + 9 x 2015 - 2015 x 6 - 2015 x 1
= 2015 x (8 + 9 - 6 - 1)
= 2015 x 10
= 20150
2015 x 8 + 9 x 2015 - 2015 x 6 - 2015
= 2015 x 8 + 9 x 2015 - 2015 x 6 - 2015 x 1
= 2015 x ( 8 + 9 - 6 - 1 )
= 2015 x 10
= 20150.
#Y/n
- ( 793 - 2015) + ( - 2015 - 1207)
= -793 + 2015 -2015 - 1207
= ( -793 - 1207) + (2015 -2015)
= -2000 +0
= -2000
- ( 793 - 2015 ) + ( - 2015 - 1207 )
=793+2015+(-2015)-1207
=(793+1207)-[2015+(-2015)]
=2000-0
=0
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
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
( 2014 x 2015 - 2016 ) : ( 2012 + 2013 x 2014 )
= ( 4058210 - 2016 ) : ( 2012 + 4054182 )
= 4056194 : 4056194
= 1
1-2013/2015=2/2015
1-2015/2017=2/2017
Vì 2/2015>2/2017=>2013/2015<2015/2017
1002 × 2015 -2015 - 2015
=1002 x 2015 - 2015 x1 - 2015 x 1
=2015 x (1002 - 1 -1 )
=2015 x 1000
=2015000
kb k ^ - ^
1002 x 2015 - 2015 - 2015
= ( 1002 - 1 - 1 ) x 2015
= 1000 x 2015
= 2015000
Tk mk nhé ~~!!^_^