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A = (n + 2015)(n + 2016) + n2 + n
= (n + 2015)(n + 2015 + 1) + n(n + 1)
Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2
=> (n + 2015)(n + 2015 + 1) chia hết cho 2
n(n + 1) chia hết cho 2
=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2
=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)
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D\(\frac{2013}{2014+2015}+\frac{2014}{2014+2015}\)
Vì \(\frac{2013}{2014}>\frac{2013}{204+2015}\)
và \(\frac{2014}{2015}>\frac{2014}{2014+2015}\)
nên C>D
Ủng hộ mk nha
\(\frac{2013}{2014}+\frac{2014}{2015}=1,999...\)
\(\frac{2013+2014}{2014+2015}=4029\)
nen D>C
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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
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Ta có:
\(\frac{2013}{2014}>\frac{2013}{2014+2015}\)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\)
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\)
\(\Rightarrow M>N\)
Ta có: \(N=\frac{2013+2014}{2014+2015}<1\);
\(M=\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013}{2015}+\frac{2014}{2015}=\frac{4027}{2015}>1\)
\(\Rightarrow A>B\)
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http://olm.vn/hỏi-đáp/question/85092.html, bạn ấn vào đây có bài y chang luôn !
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Ta có :
\(\frac{2014^{2015}+1}{2014^{2015}+1}\)\(=1\)
\(\frac{2014^{2014}+1}{2014^{2013}+1}\)\(>1\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
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\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}=3,00000074\)
Vì 3,00000074 > 3 nên A > 3
Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015}\) (1)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\) (2)
ộng caác bất đẳng thứa (1) và (2) vào vế với vế:
\(\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\Rightarrow A>B\)
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