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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A\text{=}\dfrac{2013.2014-1}{2013.2014}\text{=}\dfrac{2013.2014}{2013.2014}-\dfrac{1}{2013.2014}\text{=}1-\dfrac{1}{2013.2014}\)
\(B\text{=}\dfrac{2014.2015-1}{2014.2015}\text{=}\dfrac{2014.2015}{2014.2015}-\dfrac{1}{2014.2015}\text{=}1-\dfrac{1}{2014.2015}\)
\(Ta\) có : \(\dfrac{1}{2013.2014}>\dfrac{1}{2014.2015}\)
\(\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Q=\dfrac{2010+2011+2012}{2011+2012+2013}=\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
Ta có: \(\dfrac{2010}{2011+2012+2013}< \dfrac{2010}{2011}\)
\(\dfrac{2011}{2011+2012+2013}< \dfrac{2011}{2012}\)
\(\dfrac{2012}{2011< 2012< 2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(P>Q\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}=1-\dfrac{1}{31}=\dfrac{30}{31}\)
P=2014/2015=1-1/2015
mà 1/31>1/2015
nên S<P
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\dfrac{2014}{2015}+\dfrac{2015}{2014}=1-\dfrac{1}{2015}+1+\dfrac{1}{2014}=2-\dfrac{1}{2015}+\dfrac{1}{2014}>2\)(Vì \(\dfrac{1}{2014}>\dfrac{1}{2015}\))
\(\dfrac{666665}{333333}=2-\dfrac{1}{333333}< 2\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 173451/214263 = 17/21 = 0,8095
79/83 = 0,9518
Từ đó , suy ra : 0,9518 > 0,8095 ( hay 79/83 > 17/21 )
Vậy 173451/214263 > 79/83
\(\frac{173451}{214263}\)và \(\frac{79}{63}\)
= \(\frac{17}{21}\)và \(\frac{79}{63}\)
= \(\frac{51}{63}\)và \(\frac{79}{63}\)
= \(\frac{51}{63}\)< \(\frac{79}{63}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)
Vậy \(2^{332}< 3^{223}\)
1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)
\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)
Nên suy ra \(10A>10B\Rightarrow A>B\)
\(P=\dfrac{2013.2014-1007.4030}{2014^2-2011.2014}\)
\(P=\dfrac{2013.2014-1007.2.2015}{2014^2-2011.2014}\)
\(P=\dfrac{2013.2014-2014.2015}{2014^2-2011.2014}\)
\(P=\dfrac{2014.\left(2013-2015\right)}{2014.\left(2014-2011\right)}\)
\(P=-\dfrac{2}{3}\)
\(Q=-\dfrac{214263}{142862}=-\dfrac{214263:71421}{142862:71421}=-\dfrac{3}{2}\)
Vì \(-\dfrac{2}{3}>-\dfrac{3}{2}\)nên P>Q
\(P=\dfrac{2013.2014-1007.4030}{2014^2-2011.2014}\)
\(P=\dfrac{2013.2014-1007.4030}{2014.2014-2011.2014}\)
\(P=\dfrac{2013.2.1007-1007.4030}{2014\left(2014-2011\right)}\)
\(P=\dfrac{4026.1007-1007.4030}{2014.3}\)
\(=\dfrac{1007.\left(4026-4030\right)}{1007.2.3}\)
\(=\dfrac{1007.-4}{1007.6}=\dfrac{-4}{6}=\dfrac{-2}{3}\)
\(Q=\dfrac{214263}{142842}=-\dfrac{3}{2}\)
\(-\dfrac{3}{2}< -\dfrac{2}{3}\Rightarrow P>Q\)