a.(-2013).2014+1007.26

=(-2013).2.1007+1007.26

=(-4026).1007+1007.26

=1007.(-4026+26)

=1007.(-4000)

=-4028000

b.(1313/1414 + 10/160) - (130/140 - 1515/1616)

=(13/14+1/16)-(13/140-(1/16+15/16)

=13/14+1/16-13/14+15/16

=(13/14+13/14)-(1/16+15/16)

=26/14-16/16

=26/14-1

=26/14-14/14

=12/14

=6/7

ấn đúng cho mình nhé

kết quả là 6/7

:)
 

Ta có:

A= 1+2-3-4+5+6-7-8+...-2011-2012+2013+2014 

= (1+2-3-4)+(5+6-7-8)+...(2009+2010-2011-2012)+(2013+2014)

Ta thấy từ 1 đến 2012 có: \(x = {2012-1 \over 1}\)+1=2012(số)

Ta nhóm các số hạng kia trong tổng A và bớt đi tổng 2013+2014, mỗi nhóm là 4 số hạng liên tiếp

=> Có số nhóm là: 2012:4=503(nhóm)

Ta lại có: 

A= (1+2-3-4)+(5+6-7-8)+...(2009+2010-2011-2012)+(2013+2014)

=(-4)+(-4)+...+(-4)+(2013+2014)

(503 số hạng -4)

=(-4).503+(2013+2014)

=(-2012)+4027

=2015

Vậy A=2015

Ta có : 1+2-3-4+5+6-7-8+...-2011-2012+2013+2014
=(1+2)+(-3-4+5+6)+(-7-8+9+10)+...+(-2011-2012+2013+2014)
=3+(4+4+...+4)(có 503 số 4)
=3+4*503
=3+2012
=2015

= -2013 x 1007 x 2 + 1007 x 26

= 1007 x ( -2013 x 2 + 26 ) 

= 1007 x ( -4000) 

= -4028000

\(-2013\times2014+1007\times26\)

\(=-2013\times2014+1007\times2\times13\)

\(=(-2013+13)\times2014\)

\(=-2000\times2014\)

\(=-2\times1000\times2014\)

\(=-2\times2014000\)

\(=-4028000\)

d) 0

\(B=2^{2016}-2^{2015}+2^{2014}-2^{2013}+.......+2^2-2^1+2^0\)

\(\Rightarrow2B=2^{2017}-2^{2016}+2^{2015}-2^{2014}+.........+2^3-2^2+2^1\)

\(\Rightarrow2B+B=3B=2^{2017}+2^0\)

\(\Rightarrow B=\frac{2^{2017}+2^0}{3}=\frac{2^{2017}+1}{3}\)

a: \(5\cdot\frac{5}{27}+\frac{7}{23}+0,5\cdot\frac57+\frac{16}{23}+\frac12\)

\(=5\cdot\frac{5}{27}+\left(\frac{7}{23}+\frac{16}{23}\right)+0,5\left(\frac57+1\right)\)

\(=\frac{25}{27}+1+0,5\cdot\frac{12}{7}=\frac{25}{27}+1+\frac67=\frac{25}{27}+\frac{13}{7}=\frac{526}{27\cdot7}=\frac{526}{189}\)

b: \(25\frac16:\frac{-4}{5}-45\frac16:\frac{-4}{5}\)

\(=\left(25+\frac16-45-\frac16\right):\frac{-4}{5}\)

\(=-20\cdot\frac{-5}{4}=25\)

c: \(25\cdot\left(-\frac15\right)^3+\frac15-2\cdot\left(-\frac12\right)^2-\frac12\)

\(=25\cdot\frac{-1}{125}+\frac15-2\cdot\frac14-\frac12\)

\(=-\frac15+\frac15-\frac12-\frac12=-1\)

a: \(-\left(251\cdot3+281\right)+3\cdot251-\left(1-281\right)\)

\(=-251\cdot3-281+3\cdot251-1+281\)

=-1

b: \(\left(0,75-\frac14\right):\left(-5\right)+\frac{1}{15}-\left(-\frac15\right):\left(-3\right)\)

\(=\left(0,75-0,25\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)

=0,5:(-5)

=-1

c: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac25\)

\(=\frac29:\frac59-\frac75\)

\(=\frac25-\frac75=-\frac55=-1\)

a: \(-\left(251\cdot3+281\right)+3\cdot251-\left(1-281\right)\)

\(=-251\cdot3-281+3\cdot251-1+281\)

=-1

b: \(\left(0,75-\frac14\right):\left(-5\right)+\frac{1}{15}-\left(-\frac15\right):\left(-3\right)\)

\(=\left(0,75-0,25\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)

=0,5:(-5)

=-1

c: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac25\)

\(=\frac29:\frac59-\frac75\)

\(=\frac25-\frac75=-\frac55=-1\)

a: \(5\cdot\frac{5}{27}+\frac{7}{23}+0,5\cdot\frac57+\frac{16}{23}+\frac12\)

\(=5\cdot\frac{5}{27}+\left(\frac{7}{23}+\frac{16}{23}\right)+0,5\left(\frac57+1\right)\)

\(=\frac{25}{27}+1+0,5\cdot\frac{12}{7}=\frac{25}{27}+1+\frac67=\frac{25}{27}+\frac{13}{7}=\frac{526}{27\cdot7}=\frac{526}{189}\)

b: \(25\frac16:\frac{-4}{5}-45\frac16:\frac{-4}{5}\)

\(=\left(25+\frac16-45-\frac16\right):\frac{-4}{5}\)

\(=-20\cdot\frac{-5}{4}=25\)

c: \(25\cdot\left(-\frac15\right)^3+\frac15-2\cdot\left(-\frac12\right)^2-\frac12\)

\(=25\cdot\frac{-1}{125}+\frac15-2\cdot\frac14-\frac12\)

\(=-\frac15+\frac15-\frac12-\frac12=-1\)