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\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta thấy: \(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\)
\(\Rightarrow M=\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}>N=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Vậy M>N
\(TA-CO':\)
\(A=\frac{4+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}{7+\frac{7}{2014}-\frac{7}{2015}+\frac{7}{2012}-\frac{7}{2013}}\)
\(A=\frac{4\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}{7\left(\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4}{7}\)
\(B=\frac{1+2+...+2^{2013}}{2^{2015}-2}\)
ĐẶT \(C=1+2+...+2^{2013}\)
\(\Rightarrow2C=2+2^2+...+2^{2014}\)
\(\Rightarrow2C-C=\left(2+2^2+...+2^{2014}\right)-\left(1+2+...+2^{2013}\right)\)
\(\Rightarrow C=2^{2014}-2\)
\(\Rightarrow B=\frac{2^{2014}-1}{2^{2015}-2}\)
\(B=\frac{2^{2014}-1}{2\left(2^{2014}-1\right)}\)
\(B=\frac{1}{2}\)
\(\Rightarrow A-B=\frac{3}{7}-\frac{1}{2}=\frac{6}{14}-\frac{7}{14}\)
\(A-B=\frac{6-7}{14}=\frac{-1}{14}\)
VẬY, \(A-B=\frac{-1}{14}\)
Xét N có:
\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)
Ta các số hạng của M và N có:
\(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\) (1)
\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\) (2)
\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\) (3)
Từ (1);(2);(3) => M > N
Lời giải:
a)
\(\frac{2012}{2013}=1-\frac{1}{2013}; \frac{2013}{2014}=1-\frac{1}{2014}\)
Mà \(\frac{1}{2013}>\frac{1}{2014}\Rightarrow 1-\frac{1}{2013}< 1-\frac{1}{2014}\Rightarrow \frac{2012}{2013}< \frac{2013}{2014}\)
b)
\(\frac{1006}{1007}=1-\frac{1}{1007}\)
\(\frac{2013}{2015}=1-\frac{2}{2015}>1-\frac{2}{2014}=1-\frac{1}{1007}\)
Do đó: \(\frac{2013}{2015}> \frac{1006}{1007}\)
ta thấy:
\(\frac{2012}{2013}+\frac{2013}{2014}>\frac{2012}{2014}+\frac{2013}{2014}=\frac{2012+2013}{2014}>\frac{2012+2013}{2013+2014}\)
A= 4,00000148 nên A>4
A = 2013/2013 - 1/2013 + 2014/2014 -1/2014 + 2015/2015 - 1/2015 + 2012/2012 + 3/2012
A = 1 - 1/2013 + 1 - 1/2014 + 1 - 1/2015 + 1 = 1/2012 + 1/2012 + 1/2012
A = 4 + ( 1/2012 - 1/2013) + (1/2012 - 1/2014) + (1/2012 - 1/ 2015)
Vì:
1/2012 > 1/2013 => 1/2012 - 1/2013>0
1/2012 > 1/2014 => 1/2012 - 1/2014>0
1/2012 > 1/2015 => 1/2012 - 1/2015>0
=>( 1/2012 - 1/2013) + (1/2012 - 1/2014) + (1/2012 - 1/ 2015) >0.
=>4 + ( 1/2012 - 1/2013) + (1/2012 - 1/2014) + (1/2012 - 1/ 2015) > )+ 4 = 4.