Tính bằng cách thuận tiện nhất
f) 2015 x 8 + 7 x 2015 – 2015 x 5g) 44953 – 79 – 953 + 9
Tính bằng cách thuận tiện :
2015 x 8 + 9 x 2015 - 2015 x 6 - 2015
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
các bạn ơi toàn cách giải lớp 4 đúng không .
Tính bằng cách thuận tiện nhất:
a)13x15-150+97x15
b)2016/2015 x 4/7 -4/7 x1/2015 +3/7
Giúp mình với ạ
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
Tính bằng cách thuận tiện nhất:
a)13x15-150+97x15
b)2016/2015 x 4/7 -4/7 x1/2015 +3/7
Giúp mình với ạ
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
tính bằng cách thuận tiện nếu có thể: ( 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 ( 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
Tính bằng cách thuận tiện nhất :
2016 x 2016 + 2 / 2015 x 2016 + 2018
CHO MÌNH HỎI CÁI DẤU KIA LÀ GÌ VẬY BẠN
Tính bằng cách thuận tiện nhất : ( 2014 x 2015 - 2016 ) : (2012 + 2013 x 2014 )
( 2014 x 2015 - 2016 ) : ( 2012 + 2013 x 2014 )
= ( 4058210 - 2016 ) : ( 2012 + 4054182 )
= 4056194 : 4056194
= 1
﴾ 2014 x 2015 ‐ 2016 ﴿ : ﴾ 2012 + 2013 x 2014 ﴿
= ﴾ 4058210 ‐ 2016 ﴿ : ﴾ 2012 + 4054182 ﴿
= 4056194 : 4056194
= 1
( 2014 x 2015 - 2016 ) : (2012 + 2013 x 2014 )
= ( 2014 x ( 2013 + 2 ) - 2016 ) : ( 2012 + 2013 x 2014 )
= ( 2014 x 2013 + 2014 x 2 - 2016 ) : ( 2012 + 2013 x 2014 )
= ( 2014 x 2 - 2016 ) : 2012
= ( 4028 - 2016 ) : 2012
= 2012 : 2012
= 1
tính bằng cách thuận tiện nhất 1002 × 2015 -2015 -2015
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 ~~!!^_^
1002 x 2015 - 2015 - 2015 = 1002 x 2015 - 2015 x 1 - 2015 x 1 = 2015 x ( 1002 - 1 - 1 ) = 2015 x 1000 = 2015000
Tính bằng cách thuận tiện
- ( 793 - 2015 ) + ( - 2015 - 1207 )
- ( 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
Tính bằng cách thuận tiện nhất
2014 x 512 +2015 x 488
2014 x 512 + 2015 x 488
= 2014 x 512 + ( 2014 + 1 ) x 488
= 2014 x 512 + 2014 x 488 + 488
= 2014 x ( 512 + 488 ) + 488
= 2014 x 1000 + 488
= 2 014 000 + 488
= 2 014 488
\(\text{2014 x 512 + 2015 x 488 = 2014 x 512 + ( 2014 + 1 ) x 488 = 2014 x 512 + 2014 x 488 + 488 = 2014 x ( 512 + 488 ) + 488 = 2014 x 1000 + 488 = 2 014 000 + 488 = 2 014 488}\)
\(2014\times512+2015\times488=2014\times512+\left(2014+1\right)\times488\)
\(=2014\times512+2014\times488+488\)
\(=2014\times\left(512+488\right)+488\)
\(=2014\times1000+488\)
\(=2014000+488\)
\(=2014488\)