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\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
\(A=\frac{2015^{2013}+1}{2015^{2014}+1}=\frac{\left(2015^{2013}+1\right)\left(2015^{2014}+1\right)}{\left(2015^{2014}+1\right)\left(2015^{2016}+1\right)}=\frac{2015^{4027}+2015^{2013}+2015^{2014}+1}{\left(2015^{2014}+1\right)\left(2015^{2016}+1\right)}\)
\(B=\frac{2015^{2015}+1}{2015^{2016}+1}=\frac{\left(2015^{2015}+1\right)\left(2015^{2014}+1\right)}{\left(2015^{2016}+1\right)\left(2015^{2014}+1\right)}=\frac{2015^{4029}+2015^{2015}+2015^{2014}+1}{\left(2015^{2016}+1\right)\left(2015^{2014}+1\right)}\)
Ta thấy hiển nhiên thử của B > tử của A nên B > A
Vậy...
\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{97\cdot99}-\frac{5}{4}\cdot\frac{13}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right)-\frac{13}{4}\cdot\frac{5}{99}+\frac{5}{99}\cdot\frac{1}{4}\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{5}{99}\cdot\left(\frac{13}{4}-\frac{1}{4}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)-\frac{5}{99}\cdot3\)
\(A=\frac{1}{2}\cdot\frac{32}{99}-\frac{5}{33}\)
\(A=\frac{16}{99}-\frac{5}{33}=\frac{1}{99}\)
\(9A=\frac{9\left(9^{2014}+1\right)}{9^{2015+1}}=\frac{9^{2015}+9}{9^{2015}+1}=\frac{9^{2015}+1+8}{9^{2015}+1}=1+\frac{8}{9^{2015}+1}\)
\(9B=\frac{9\left(9^{2015}+1\right)}{9^{2016+1}}=\frac{9^{2016}+9}{9^{2016}+1}=\frac{9^{2016}+1+8}{9^{2016}+1}=1+\frac{8}{9^{2016}+1}\)
Ta thấy \(9^{2016}+1>9^{2015}+1\Rightarrow\frac{8}{9^{2016}+1}<\frac{8}{9^{2015}+1}\)
suy ra 9A >9B
Vậy A > B
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