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\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
Ta có:
A = -1 – 2 + 3 + 4 – 5 – 6 + 7 + 8 – 9 – 10 + 11 + 12 - ... – 2013 – 2014 + 2015 + 2016
A = (0 – 1 – 2 + 3) + (4 – 5 – 6 + 7) + ... + (2012- 2013 – 2014 + 2015) + 2016
A = 0 + 0 + ... + 0
A = 2016
Vậy A = 2016
#Mạt Mạt#
A = (-1 + -5 + -9 +...... + -20130) + (-2 + -6 + -10 + ... + -2014) + ( 3 + 7 + 11 +... + 2015) + (4 + 8 + 12 +.... +2016)
A = -507528 + -508032 + 508032 + 509040
A = 1512
:)))
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
A=(0-1-2+3)+(4-5-6+7)+...+((2012-2013-2014+2015)+2016
A=0+0+...+0
A=2016
Vậy A=2016
Ta có:
A = -1 – 2 + 3 + 4 – 5 – 6 + 7 + 8 – 9 – 10 + 11 + 12 - ... – 2013 – 2014 + 2015 + 2016
A = (0 – 1 – 2 + 3) + (4 – 5 – 6 + 7) + ... + (2012- 2013 – 2014 + 2015) + 2016
A = 0 + 0 + ... + 0
A = 2016
Vậy A = 2016
#Mạt Mạt#
A=-1+(-2+3+4-5)+(-6+7+8-9)+(-10+11+12-13)+.......+(-2010+2011+2012-2013)+(-2014+2015+2016)
A=-1+0+0+0+...+0+2017
A=2018
A = -1 - 2 + 3 + 4 - 5 - 6 + 7 + 8 - 9 - 10 + 11 + 12 - ... - 2013 - 2014 + 2015 + 2016
=> A = (-1 - 2 + 3 + 4) + (-5 - 6 + 7 + 8) + (-9 - 10 + 11 + 12) + ... + (-2013 - 2014 + 2015 + 2016)
=> A = 4 + 4 + 4 + ... + 4
có 2016 : 4 = 504 (số 4)
=> A = 4 . 504
=> A = 2016
= (1/2013 + 12/2014 + 123/2015) x 0
= 0