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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 \(2015^{2016}>2016^{2015}.\)
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 = \(\frac{5^{2017}-5}{4}\)
a, Ta có: \(\left(2x+\dfrac{1}{4}\right)^4\ge0\rightarrow\left(2x+\dfrac{1}{4}\right)^4+6\ge6\)
Dấu ''=" xảy ra khi \(2x+\dfrac{1}{4}=0\rightarrow2x=\dfrac{-1}{4}\rightarrow x=\dfrac{-1}{8}\)
Vậy MinE=6\(\Leftrightarrow x=\dfrac{-1}{8}\)
b, Ta có: \(\left(5-3x\right)^2\ge0\rightarrow\left(5-3x\right)^2-2013\ge-2013\)
Dấu ''='' xảy ra khi \(5-3x=0\rightarrow3x=5\rightarrow x=\dfrac{5}{3}\)
Vậy MinE=-2013\(\Leftrightarrow x=\dfrac{5}{3}\)
a) \(E=\left(2x+\dfrac{1}{4}\right)^4+6\)
Vì \(\left(2x+\dfrac{1}{4}\right)^4\ge0\)
Nên \(\left(2x+\dfrac{1}{4}\right)^4+6\ge6\)
Vậy GTNN của \(E=6\) khi \(2x+\dfrac{1}{4}=0\Leftrightarrow x=\dfrac{-1}{8}\)
b) \(E=\left(5-3x\right)^2-2013\)
Vì \(\left(5-3x\right)^2\ge0\)
Nên \(\left(5-3x\right)^2-2013\ge-2013\)
Vậy GTNN của \(E=-2013\) khi \(5-3x=0\Leftrightarrow x=\dfrac{5}{3}\)
c) \(A=2013+\left|2x-3\right|\)
Vì \(\left|2x-3\right|\ge0\)
Nên \(2013+\left|2x-3\right|\ge2013\)
Vậy GTNN của \(A=2013\) khi \(2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
d) \(B=-1+\left|\dfrac{1}{2}x-3\right|\)
Vì \(\left|\dfrac{1}{2}x-3\right|\ge0\)
Nên \(-1+\left|\dfrac{1}{2}x-3\right|\ge-1\)
Vậy GTNN của \(B=-1\) khi \(\dfrac{1}{2}x-3=0\Leftrightarrow x=6\)
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(\Leftrightarrow m=5\)
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow n=3\)