Tính bằng cách thuận tiện
1/1+2 + 1/1+2+3 + 1/1+2+3+4 + ... + 1/1+2+3+...+2014 + 1/1+2+...+2015
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 hợp lý
A=2016/1+2015/2+2014/3+...+2/2015+1/2016
Tính bằng cách thuận tiện nhất:
2015/2019 : 1/2 + 3/2019 :1/2 + 1/2019 : 1/2
=(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
2015/2019 : 1/2 + 3/2019 : 1/2 + 1/2019 : 1/2
= ( 2015/2019 + 3/2019 + 1/2019 ) : 1/2
= 1 : 1/2
= 2
Chúc bn hok tốt ! ^_^
1*2*3*...*2015-1*2*3*4*...*2014-1*2*3*...*2013*2014^2 tính nhanh
\(1.2.3....2015-1.2.3....2014-1.2.3....2013.2014^2\)
\(=1.2.3...\left(2014+1\right)-1.2.3...\left(2014+1\right)\)
\(=0\)
bạn làm thế là sai rồi
có 3 con 2014 cơ mà
Cho A= 1/2+1/3+1/4+..+1/2016
B= 2015/1+2014/2+2013/3+....+2/2014+1/2015. Tính B/A
Tính: (1*2015+2*2014+3*2013+...+2015*1)/(1*2+2*3+3*4+4*5+...+2015*2016)
Tính M=1/1*2*3+1/2*3*4+...1/2014*2015*2016.
Nhớ ghi cách trình bày giúp mik nha~
2M=2/1*2*3+...+2/2014*2015*2016
=1/1*2-1/2*3+...+1/2014*2015-1/2015*2016
=1/1*2-1/2015*2016(khử liên tiếp)
=1/2-1/4062240=2031119/4062240.
tính bằng cách thuận tiện
1/2:3/4+5/6:3/4-2/8:3/4
`1/2:3/4+5/6:3/4-2/8:3/4`
`=1/2 xx 4/3 + 5/6 xx 4/3 - 2/8 xx 4/3`
`=4/3 xx ( 1/2 + 5/6 - 2/8)`
`=4/3 xx ( 1/2 + 5/6 - 1/4 )`
`=4/3 xx ( 6/12 + 10/12 - 3/12)`
`=4/3 xx ( 16/12 - 3/12)`
`=4/3 xx 13/12`
`=13/9`
Cho A = 1/2 + 1/3 + 1/4 + ...+ 1/2016
B = 1/2015 + 2/2014 + 3/2013 + ... + 2014/2 + 2015/1
Tính B : A
\(B=\left(\dfrac{1}{2015}+1\right)+\left(\dfrac{2}{2014}+1\right)+\left(\dfrac{3}{2013}+1\right)+...+\left(\dfrac{2014}{2}+1\right)+1\)
\(=\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2016}\)
=>B:A=2016