Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c=1+2-3-4+5+6-7-.......+2014-2015-2016+2017+2018
c=-4+-4+.....+-4+-4+2018
C=(-4).1009+2018\
C=-4036+2018
c=-2018
1)
a) - 37 + 54 + (- 70) + (-163) + 246
= (-37 + (-163)) + (54 + 246) + (-163)
= -200 + 300 + (- 163)
= 100 + (-163)
= - 63
b) 125. (-61) . 23. (-1)2n
= (125 . 23) . (-61) . 1
= 1000 . (- 61)
= - 6100
c) mk không biết làm nha
a)-37+54+-70+-163+246
=[(-370 + (-70)+(-163)]+(54 + 246)
=(-603)+300
=-303
c)1+2-3-4+5+6 -7....+2014-2015-2016+2017+2018
=-4+-4+......+-4+2018
=(-4).505+2018
=-2020+2018
=-2
https://olm.vn/hoi-dap/detail/104380939254.html?pos=228315034049
bạn coi thử nha
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) 5^2018:2^2015-6^2+2017^0
=5^3-36+1
=125-36+1
=89+1=90
b)12:{390:[500-(5^3+35.7)]}
=12:{390:[500-(125+245)]}
=12:{390:[500-370]}
=12:{390:130}
=12:3=4
Hok tốt
A =\(\frac{2}{7}\times\frac{7}{15}-\frac{2}{3}\times\frac{5}{27}\) =\(\frac{14}{105}-\frac{10}{81}\)=\(\frac{2}{15}-\frac{10}{81}\)=\(\frac{162}{1215}-\frac{150}{1215}\)=\(\frac{12}{1215}\)
\(\frac{-4}{9}\times\frac{7}{15}+\frac{4}{9}\times\frac{5}{27}\)=\(\frac{-28}{135}+\frac{20}{243}\)=\(\frac{-6804}{32805}+\frac{2700}{32805}\)=\(\frac{-4104}{32805}\)=\(\frac{-152}{1215}\)
A=\(\frac{2}{7}\times\frac{7}{15}-\frac{2}{3}\times\frac{5}{27}\)chia cho\(\frac{-4}{9}\times\frac{7}{15}+\frac{4}{9}\times\frac{5}{27}\)
= \(\frac{12}{1215}:\frac{-152}{1215}\)
=\(\frac{12}{1215}\times\frac{1215}{-152}\)
=\(\frac{14580}{-184680}\)
\(\frac{14580}{-184680}\)rút gọn bằng\(\frac{-3}{38}\)
\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)
\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(=\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{2}{9}+...+\dfrac{2}{2015}-\dfrac{2}{2017}\)
\(=\dfrac{2}{3}-\dfrac{2}{2017}\)
\(=\dfrac{4028}{6051}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=\(\dfrac{1}{3}-\dfrac{1}{2017}\)
= \(\dfrac{2014}{6051}\)