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Áp dung công thức \(a>b\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}>\frac{10^{2017}+1+9}{10^{2016}+1+9}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2015}+1\right)}=\frac{10^{2016}+1}{10^{2015}+1}=A\)
\(\Leftrightarrow B>A\)
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
\(A=\frac{2015^{2013}+1}{2015^{2014}+1}=\frac{\left(2015^{2013}+1\right)\left(2015^{2014}+1\right)}{\left(2015^{2014}+1\right)\left(2015^{2016}+1\right)}=\frac{2015^{4027}+2015^{2013}+2015^{2014}+1}{\left(2015^{2014}+1\right)\left(2015^{2016}+1\right)}\)
\(B=\frac{2015^{2015}+1}{2015^{2016}+1}=\frac{\left(2015^{2015}+1\right)\left(2015^{2014}+1\right)}{\left(2015^{2016}+1\right)\left(2015^{2014}+1\right)}=\frac{2015^{4029}+2015^{2015}+2015^{2014}+1}{\left(2015^{2016}+1\right)\left(2015^{2014}+1\right)}\)
Ta thấy hiển nhiên thử của B > tử của A nên B > A
Vậy...
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
1/ ta có:
A = \(\frac{10^{2015}+1}{10^{2016}+1}\Rightarrow10A=\frac{10^{2016}+10}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
B = \(\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10B=\frac{10^{2017}+10}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
vì \(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\) => 10A > 10B
=> A > B
vậy A > B
2/ ta có: M = 5 + 52 + 53 + ... + 52016
=> 5M = 52+53+54+...+52017
=> 5M - M = (52+53+54+...+52017) - (5+52+53+...+52016)
=> 4M = 52017- 5
=> M = \(\frac{5^{2017}-5}{4}\)
vậy M = \(\frac{5^{2017}-5}{4}\)
là 142015 =10452016