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So sánh
M = \(\dfrac{2017^{2015}+1}{2017^{2015}-1}\) và N = \(\dfrac{2017^{2015}-5}{2017^{2015}-3}\)
Ta có:
M=\(\dfrac{2017^{2015}+1}{2017^{2015}-1}=\dfrac{2017^{2015}-1+2}{2017^{2015}-1}=1+\dfrac{2}{2017^{2015}-1}>1\left(1\right)\)
N=\(\dfrac{2017^{2015}-5}{2017^{2015}-3}=\dfrac{2017^{2015}-3-2}{2017^{2015}-3}=1-\dfrac{2}{2017^{2015}-3}< 1\left(2\right)\)
Từ (1) và (2) suy ra M>1>N
Vậy M>N.
Ta có :
\(\dfrac{2017^{2015}+1}{2017^{2015}-1}>\dfrac{2017^{2015}}{2017^{2015}}>\dfrac{2017^{2015}-5}{2017^{2015}-3}\)
Tick mình nha bạn hiền.
Ta có :
\(N=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}+\dfrac{-8}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2006}}\)
\(M=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-8}{10^{2005}}+\dfrac{-7}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2005}}\)
Lại có :
\(-\dfrac{8}{10^{2006}}>\dfrac{-8}{10^{2005}}\Leftrightarrow M>N\)
\(B=\frac{215-2}{2015^m}+\frac{2015+2}{2015^n}=\frac{2015}{2015^m}-\frac{2}{2015^m}+\frac{2015}{2015^n}+\frac{2}{2015^n}=A-2\left(\frac{1}{2015^m}-\frac{1}{2015^n}\right)\)
+ Nếu \(m>n\Rightarrow2015^m>2015^n\Rightarrow\frac{2}{2015^m}<\frac{2}{2015^n}\Rightarrow\frac{2}{2015^m}-\frac{2}{2015^n}<0\Rightarrow A-\left(\frac{2}{2015^m}-\frac{2}{2015^n}\right)>A\)
=> A<B
+ Nếu
m<n làm tương tự => A>B
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
\(\dfrac{2013}{2013+2014}< \dfrac{2013}{2013+2013}=\dfrac{1}{2}\)
Tương tự cộng theo vế suy ra đpcm
Ta có : A= \(\dfrac{1}{1^m}\) +\(\dfrac{1}{1^n}\)
Và B=\(\dfrac{2015-2}{2015^m}+\dfrac{2015+2}{2015^n}\)
\(\Rightarrow\)\(\dfrac{1}{m}-\dfrac{2}{2015^m}+\dfrac{1}{n}+\dfrac{2}{2015^n}\)
\(\Rightarrow\dfrac{1}{n}+\dfrac{1}{m}+\dfrac{2\left(n-m\right)}{2015^{mn}}\)
TH1 2(n-m) >0 \(\Rightarrow\) 2015mn >0 \(\Rightarrow\) A>B
TH2 2(n-m)<0\(\Rightarrow\) 2015mn<0\(\Rightarrow\) A<B
TH3 2(n-m)=0\(\Rightarrow\) 2015mn=0 \(\Rightarrow\) A=B
Xong rồi nấm ơi, bảo uyên nữa nhé