Cho A=\(\frac{2014}{2015}+\frac{2015}{2016}\)
B=\(\frac{2014+2015}{2016+2015}\)
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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)
Cho A= \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}.\). So sánh A với 4
\(A=\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2014}\\ =1-\dfrac{1}{2015}+1-\dfrac{1}{2016}+1-\dfrac{1}{2017}+1+\dfrac{1}{2014}+\dfrac{1}{2014}+\dfrac{1}{2014}\\ =\left(1+1+1+1\right)+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\\ =4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\)
Vì \(\dfrac{1}{2015}< \dfrac{1}{2014}\), \(\dfrac{1}{2016}< \dfrac{1}{2014}\), \(\dfrac{1}{2017}< \dfrac{1}{2014}\)
\(\Rightarrow\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)< 0\\ \Rightarrow-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\\>0\\ \Rightarrow4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]>4\)
Tạm thời chỉ nghĩ ra được cách này -_-
Ta có :
\(A=\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}\)
\(A=\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2014+2}{2014}\)
\(A=\frac{2015}{2015}-\frac{1}{2015}+\frac{2016}{2016}-\frac{1}{2016}+\frac{2014}{2014}+\frac{2}{2014}\)
\(A=1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{2}{2014}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2015}+\frac{1}{2016}-\frac{2}{2014}\right)\)
\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]\)
Lại có :
\(\frac{1}{2015}< \frac{1}{2014}\)
\(\frac{1}{2016}< \frac{1}{2014}\)
\(\Rightarrow\)\(\frac{1}{2015}+\frac{1}{2016}< \frac{1}{2014}+\frac{1}{2014}\)
\(\Rightarrow\)\(\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)< 0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]>3\)
Vậy \(A>3\)
Chúc bạn học tốt ~
\(y=\frac{2014}{\frac{2015}{\frac{2015}{2016}}}=\frac{2014}{2015}.\frac{2015}{2016}=\frac{1007}{1008}=1-\frac{1}{2008}\)
\(\frac{2014}{2015}=1-\frac{1}{2015}\)
Vì \(\frac{1}{2008}>\frac{1}{2015}\)nên \(\frac{1007}{1008}< \frac{2014}{2015}\)
Vậy A>y
A = \(\frac{2015^{2016}+1}{2015^{2015}+1}=\frac{2015^{2015}+1}{2015^{2015}+1}+\frac{2015}{2015^{2015}+1}=1+\frac{2015}{2015^{2015}+1}\)
B = \(\frac{2014^{2015}+1}{2014^{2014}+1}=\frac{2014^{2014}+1}{2014^{2014}+1}+\frac{2014}{2014^{2014}+1}=1+\frac{2014}{2014^{2014}+1}\)
Rồi bạn tự so sánh nha
Ta có: 2014/2015>2014/(2015+2016)
2015/2016>2015/(2015+2016)
=>2014/2015+2015/2016>2014/(2015+2016)+2015/(2015+2016)
hay 2014/2015+2015/2016>(2014+2015)/(2015+2016)
hay A>B
Vậy A>B
Đề là j vậy bạn?