Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1
Ta có : 2016/2017<1
2017/2018<1
Nên 2016/2017=2017/2018
Bài 1 :
a) Ta có : \(\frac{2016}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
Vì \(-\frac{1}{2017}< -\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\)
b) Ta có : \(\frac{2018}{2017}=1+\frac{1}{2017}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{2017}< \frac{1}{2016}\) nên \(\frac{2018}{2017}< \frac{2017}{2016}\)
Câu 2 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
\(\frac{2016\times2018+2}{2016\times2017+2018}=\frac{2016\times\left(2017+1\right)+2}{2016\times2017+2018}=\)\(=\frac{2016\times2017+2016+2}{2016\times2017+2018}=\frac{2016\times2017+2018}{2016\times2017+2018}=1\)
\(\frac{2016}{2017}< \frac{2017}{2018}\)
Đúng 100%
Đúng 100%
Đúng 100%
2012+2013.2014
2014.2015-2016
2012+2013.2014
2014.(2013+2)-2016
2012+2013.2014
2014.2013+4028-2016
2012
2012
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2015}{2016}\)
\(=\frac{1.2.3.4...2015}{2.3.4.5...2016}\)
\(=\frac{1}{2016}\)
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
\(\frac{1}{2016}\)+ \(\frac{3}{2016}\)+ \(\frac{5}{2016}\)+..........+ \(\frac{2015}{2016}\)= \(\frac{1+3+5+....+2015}{2016}\)
=\(\frac{1016064}{2016}\)= \(504\)
\(\frac{1}{2016}\)\(+\frac{3}{2016}\)\(+\frac{5}{2016}\)\(+...+\frac{2015}{2016}\)
\(=\frac{1+3+5+...+2015}{2016}\)
\(=\frac{1016064}{2016}\)
\(=504\)