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\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(1-\frac{1}{2015}\right)+\left(1-\frac{1}{2016}\right)+\left(1+\frac{2}{2014}\right)\)
\(=3-\left(\frac{1}{2015}-\frac{1}{2016}+\frac{2}{2014}\right)\)
Dễ thấy \(\frac{1}{2015}-\frac{1}{2016}+\frac{2}{2014}>0\) vì \(\frac{1}{2015}>\frac{1}{2016}\)
Do đó \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}< 3\)
A = 2014*2015 + 2015/2016 + 2016/2014
A = (1 - 1/2015) + (1 - 1/2016) + (1 + 2/2014)
A = 3 + (2/2014 - 1/2015 - 1/2016)
A = 3 + (2*2015*2016 - 2014*2016 - 2014*2015) / (2014*2015*2016)
Đặt B = 2*2015*2016 - 2014*2016 - 2014*2015
Ta có: A = 3 + B/(2014*2015*2016)
Nhận xét: Từ các phép biến đổi trên ta thấy A là tổng của 3 với một phân số có mẫu số dương. Do vậy, để so sánh A với 3 ta chỉ cần so sánh B với 0.
B = 2*2015*2016 - 2014*2016 - 2014*2015
B = 2016(2*2015 - 2014) - 2014*2015
B = 2016(2*2015 - 2014) - 2014(2016 - 1)
B = 2016(2*2015 - 2014) - 2014*2016 + 2014
B = 2016(2*2015 - 2014 - 2014) + 2014
B = 2016(2*2015 - 2*2014) + 2014
B = 2*2016(2015 - 2014) + 2014
B = 2*2016 + 2014 > 0
Vậy A > 3 (Đáp số)
Ta có : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)
Mà : \(\left(\frac{2014}{2015}+\frac{1}{2014}\right)>1;\left(\frac{2015}{2016}+\frac{1}{2014}\right)>1;\frac{2014}{2014}=1\)
Nên : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)\(>1+1+1=3\)
Ta có:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)
Mà:\(\left(\frac{2014}{2015}+\frac{1}{2014}\right)>1:\left(\frac{2015}{2016}+\frac{1}{2014}\right)>\)\(1:\frac{2014}{2014}=1\)
Nên:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}>1+1+1=3\)
\(\frac{2016}{2015}>1;\frac{2015}{2014}>1;\frac{2014}{2013}>1.\)
\(\Rightarrow\frac{2016}{2015}+\frac{2015}{2014}+\frac{2014}{2013}>1+1+1=3\)
Vậy A>3
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 \(2015^{2016}>2016^{2015}.\)
Khi a=1/2015 thì \(P=\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
\(=\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\)
\(=\dfrac{1}{2014}-\dfrac{1}{2016}=\dfrac{2}{2014\cdot2016}=\dfrac{1}{1008\cdot2014}\)
\(=\dfrac{1}{2030112}\)
