tính bằng cách thuận tiện nhất
2015x112-2015x11-2015
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( 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
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
=(2015/ 2019 + 3/2019 + 1/2019 ) : 1/2
= 2019/2019 x 2
= 1 x2
=2
2015/2019:1/2+3/2019:1/2+1/2019:1/2
=(2015/2019+3/2019+1/2019):1/2
=1:1/2
=2
k cho mink nha
- ( 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
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
`2015 xx 112 - 2015 xx 11 - 2015`
`=2015 xx 112 - 2015 xx 11 - 2015 xx 1`
`=2015 xx (112 - 11 - 1)`
`=2015 x 100`
`=251500`
= 2015x112 -2015x11-2015
=2015x112 -2015x11-2015x1
=2015x(112-11-1)
=2015x100
=201500