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ta có :
\(10A=\frac{10^{2014}+10}{10^{2014}+1}=\frac{\left(10^{2014}+1\right)+9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
\(10B=\frac{10^{2015}+10}{10^{2015}+1}=\frac{\left(10^{2015}+1\right)+9}{10^{2015}+1}=1+\frac{9}{10^{2015}+1}\)
ta thấy \(10^{2014}+1< 10^{2015}+1\Rightarrow\frac{9}{10^{2014}+1}>\frac{9}{10^{2015}+1}\Rightarrow10A>10B\Rightarrow A>B\)
Tổng S có 50 phân số
=> S > 1/100 + 1/100 + 1/100 +...+ 1/100 (50 phân số) => S > 1/2.
Vậy S > 1/2
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Rightarrow A>B\)
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Theo đề ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Lúc này ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Leftrightarrow A>B\).
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}\Leftrightarrow D< 1\left(đpcm\right)\)
cùng nhân tử với 2014>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>