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Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
\(B=\frac{215-2}{2015^m}+\frac{2015+2}{2015^n}=\frac{2015}{2015^m}-\frac{2}{2015^m}+\frac{2015}{2015^n}+\frac{2}{2015^n}=A-2\left(\frac{1}{2015^m}-\frac{1}{2015^n}\right)\)
+ 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
Phân tích từ B ra, ta có:
B=\(\dfrac{2015+2016}{2016+2017}\)=\(\dfrac{2015}{2016+2017}\)+\(\dfrac{2016}{2016+2017}\)
Vì \(\dfrac{2015}{2016+2017}\)<\(\dfrac{2015}{2016}\) ; \(\dfrac{2016}{2016+2017}\) < \(\dfrac{2016}{2017}\)
=> B < A
2) Để A là nguyên thì n - 1 là ước nguyên của 2
\(n-1=1\Rightarrow n=2\)
\(n-1=2\Rightarrow n=3\)
3) Ta gọi M là \(\dfrac{12}{5^{2012}}\)
\(M=\dfrac{5.12}{5^{2012}.5}=\dfrac{60}{5^{2013}}\)
\(\Rightarrow\) \(A=\dfrac{60}{5^{2013}}+\dfrac{18}{5^{2013}}=\dfrac{78}{5^{2013}}\)
Ta gọi Q là \(\dfrac{18}{5^{2012}}\)
\(Q=\dfrac{18}{5^{2012}}=\dfrac{18.5}{5^{2012}.5}=\dfrac{90}{5^{2013}}\)
\(\Rightarrow\) \(B=\dfrac{90}{5^{2013}}+\dfrac{12}{5^{2013}}=\dfrac{102}{5^{2013}}\)
\(\dfrac{90}{5^{2013}}< \dfrac{102}{5^{2013}}\Rightarrow A< B\)
Ai thấy đúng thì ủng hộ mink, thấy sai góp ý nha !!!
Ta có: \(M=\frac{2017^{2015}+1}{2017^{2015}-1}=\frac{2017^{2015}-1+2}{2017^{2015}-1}=1+\frac{2}{2017^{2015}-1}\)
\(N=\frac{2017^{2015}-5}{2017^{2015}-3}=\frac{2017^{2015}-3-2}{2017^{2015}-3}=1-\frac{2}{2017^{2015}-3}\)
Vì \(\frac{2}{2017^{2015}-1}>-\frac{2}{2017^{2015}-3}\)nên M>N
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