( 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

=(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

\(1.2.3....2015-1.2.3....2014-1.2.3....2013.2014^2\)

\(=1.2.3...\left(2014+1\right)-1.2.3...\left(2014+1\right)\)

\(=0\)

bạn làm thế là sai rồi 

có 3 con 2014 cơ mà 

Gọi phân số đã cho là \(A\).

Ta có:

\(A=\dfrac{1\times2015+2\times2014+3\times2013+\cdots+2015\times1}{1\times2+2\times3+3\times4+\cdots+2015\times2016}\)

Tử số:

Ta nhận thấy:

\(1\times2015+2\times2014+\cdots+2015\times1\)

\(=1(2016-1)+2(2016-2)+\cdots+2015(2016-2015)\)

\(=2016(1+2+\cdots+2015)-(1^2+2^2+\cdots+2015^2)\).

Mẫu số:

\(1\times2+2\times3+\cdots+2015\times2016\)

\(=1(1+1)+2(2+1)+\cdots+2015(2015+1)\)

\(=(1^2+2^2+\cdots+2015^2)+(1+2+\cdots+2015)\).

Đặt \(S=1+2+\cdots+2015\), \(T=1^2+2^2+\cdots+2015^2\).

Khi đó:

\(A=\dfrac{2016S-T}{T+S}\).

Với \(S=\dfrac{2015\times2016}{2}\),

\(T=\dfrac{2015\times2016\times4031}{6}\).

Thay vào:

\(A=\dfrac{2016\times\dfrac{2015\times2016}{2}-\dfrac{2015\times2016\times4031}{6}}{\dfrac{2015\times2016\times4031}{6}+\dfrac{2015\times2016}{2}}\)

Rút gọn \(2015\times2016\):

\(A=\dfrac{2016/2-4031/6}{4031/6+1/2}\)

\(=\dfrac{6048-4031}{4031+3}\)

\(=\dfrac{2017}{4034}=\dfrac12\).

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.

Cảm ơn bạn

`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`

\(B=\left(\dfrac{1}{2015}+1\right)+\left(\dfrac{2}{2014}+1\right)+\left(\dfrac{3}{2013}+1\right)+...+\left(\dfrac{2014}{2}+1\right)+1\)

\(=\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2016}\)

=>B:A=2016