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a)\(\frac{2016}{2017}< 1;\frac{2015}{2016}< 1\)
b)\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và
\(\frac{2016}{2017}< 1;\frac{2016}{2015}< 1\)
\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và \(\frac{2015}{2016}\)< \(\frac{2017}{2016}\)và \(\frac{2016}{2015}\)
\(x=\frac{-2014}{-2013}=\frac{2014}{2013}>1\)
\(y=\frac{-2015}{-2016}=\frac{2015}{2016}
2017/2016-1=2017/2017-2016/2017=1/2017
2016/2015-1=2016/2016-2015/2016=1/2016
Vì 1/2017<1/2016nên 2017/2016<2016/2015
Ta có :
1 - 2017 /2016 = 1/2016
1 - 2016/2015 = 1/2015
Vì 1/2015 > 1/2016
nên ..............
a/ \(3^{150}=\left(3^2\right)^{75}=9^{75}\)
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(9^{75}>8^{75}\Rightarrow3^{150}>2^{225}\)
b/
\(20162016^{10}=\left(2016.10001\right)^{10}=2016^{10}10001^{10}\)
\(2016^{20}=2016^{10}.2016^{10}\)
\(10001^{10}>2016^{10}\Rightarrow2016^{10}.10001^{10}>2016^{10}.2016^{10}\Rightarrow20162016^{10}>2016^{20}\)
c/ \(\frac{222^{333}}{333^{222}}=\frac{\left(222^3\right)^{111}}{\left(333^2\right)^{111}}=\frac{\left(2^3.111^3\right)^{111}}{\left(3^2.111^2\right)^{111}}=\left(\frac{8.111}{9}\right)^{111}\)
\(\frac{888}{9}>1\Rightarrow\left(\frac{888}{9}\right)^{111}>1\Rightarrow222^{333}>333^{222}\)
a) Ta có: 3^150 = 3^2.75 = (3^2)^75 = 9^75
2^225 = 2^3.75 = (2^3)^75 = 8^75
Vì 9 > 8 nên 9^75 > 8^75
Vậy 3^150 > 2^225
b) Ta có: 2016^20 = 2016^10+10 = 2016^10 . 2016^10
20162016^10 = (10001 . 2016)^10 = 10001^10 . 2016^10
Vì 2016^10 < 10001^10 nên 2016^10 . 2016^10 < 10001^10 . 2016^10
Vậy 2016^20 < 20162016^10
Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a \(\ne\) b ta có:
\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)
\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)
\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{2015}\)
\(\Rightarrow\sqrt{2016}+\sqrt{2014}< 2.\sqrt{2015}\)
\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)
Ta có:
\(\left(2015^{2015}+2016^{2015}\right)^{2016}=\left(2015^{2015}+2016^{2015}\right)^{2015}.\left(2015^{2015}+2016^{2015}\right)\)
\(>\left(2015^{2015}+2016^{2015}\right)^{2015}.2016^{2015}=\left[\left(2015^{2015}+2016^{2015}\right)2016\right]^{2015}\)
\(>\left(2015^{2015}.2015+2016^{2015}.2016\right)^{2015}=\left(2015^{2016}+2016^{2016}\right)^{2015}\)
Vậy \(\left(2015^{2015}+2016^{2015}\right)^{2016}>\left(2015^{2016}+2016^{2016}\right)^{2015}\)
1. Ta sẽ chứng minh \(2015^{2016}>2016^{2015}\)
\(\Leftrightarrow2016^{2015}-2015^{2016}< 0\Leftrightarrow2016^{2016}-2016.2015^{2016}< 0\)
\(\Leftrightarrow2016.2016^{2016}-2015.2016^{2016}-2016.2015^{2016}< 0\)
\(\Leftrightarrow2016\left(2016^{2016}-2015^{2016}\right)< 2015.2016^{2016}\)
\(\Leftrightarrow2016\left(2016^{2015}+2016^{2014}.2015+...+2015^{2015}\right)< 2015.2016^{2016}\)
\(\Leftrightarrow2016^{2015}.2015+...+2016.2015^{2015}< 2014.2016^{2016}\)
\(\Leftrightarrow2016^{2014}.2015+2016^{2013}.2015^2+...+2015^{2015}< 2014.2016^{2015}\)
\(\Leftrightarrow2015^{2015}< \left(2016^{2015}-2015.2016^{2014}\right)+\left(2016^{2015}-2015^2.2016^{2013}\right)\)
\(+...+\left(2016^{2015}-2015^{2014}.2016\right)\)
\(\Leftrightarrow2015^{2015}< 2014.2016^{2014}+2013.2016^{2014}.2015+...+2016.2015^{2013}\)
Lại có \(2015^{2015}=2014.2015^{2014}+2015^{2014}< 2014.2016^{2014}+2015^{2014}\)
Mà \(2015^{2014}< 2013.2016^{2014}.2015\)
nên \(2015^{2014}< 2014.2016^{2014}+2013.2016^{2014}.2015+...+2016.2015^{2013}\)
Vậy \(2015^{2016}>2016^{2015}.\)