\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)

\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)

\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)

\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)

( 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

Đề có sai kh vậy bạn

Sao tui thấy nó cứ sai sai 

Bạn xem lại đề nhá !!!

#hoc_tot#

C = 1 - 2 - 3 + 4 + 5 - 6 - 7 - 8 + ... + 2013 - 2014 - 2015 + 2016 + 2017 - 2018

Số số hạng của C từ 1 đến 2016 là : ( 2016 - 1 ) : 1 + 1 = 2016 số

Nhóm 4 số thành 1 cặp ta có : 2016 : 4 = 504 cặp

=> C = ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ... + ( 2013 - 2014 - 2015 + 2016 ) + 2017 - 2018

C = 0 + 0 + ... + 0 + ( -1 )

C = -1

 ๖ۣۜHσàηɠ ︵ɱ√ρ Đề k sai ... Đơn giản vì bạn k bt làm =))

\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)

\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)

\(B=\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+,...+\left(-2014\right)+2016\)

\(B=2+2+....+2\left(\text{504 số hạng 2}\right)=1008\)

c) 1 + ( -3 ) +5 + ( -7 ) + ...........+ 2013 + ( -2015 )

[ 1 + (-3 ) ] + [ 5 + -7 ] + .......... + [ 2013 + ( - 2015 ) ]

 có số cặp là : [ ( 2015 - 1 ) : 2 + 1 ] : 2 = 504 ( cặp )

= -2 + -2 + -2 +..........+ -2

= -2 x 504

= -1008

Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)

\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)

\(=2017\)

nhieu qua zay bn

ai ma tinh duoc

ha bn?