Cho \(A=\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}\) .Hãy so sánh A với 3
K
Khách
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Những câu hỏi liên quan
PA
1 tháng 9 2016
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)
NH
16 tháng 6 2016
\(y=\frac{2014}{\frac{2015}{\frac{2015}{2016}}}=\frac{2014}{2015}.\frac{2015}{2016}=\frac{1007}{1008}=1-\frac{1}{2008}\)
\(\frac{2014}{2015}=1-\frac{1}{2015}\)
Vì \(\frac{1}{2008}>\frac{1}{2015}\)nên \(\frac{1007}{1008}< \frac{2014}{2015}\)
Vậy A>y
17 tháng 7 2017
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 : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)\(>1+1+1=3\)
VT
17 tháng 7 2017
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:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}>1+1+1=3\)
LV
Cho A= \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2014}.\). So sánh A với 4
1
MV
28 tháng 4 2017
\(A=\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2014}\\ =1-\dfrac{1}{2015}+1-\dfrac{1}{2016}+1-\dfrac{1}{2017}+1+\dfrac{1}{2014}+\dfrac{1}{2014}+\dfrac{1}{2014}\\ =\left(1+1+1+1\right)+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\\ =4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\)
Vì \(\dfrac{1}{2015}< \dfrac{1}{2014}\), \(\dfrac{1}{2016}< \dfrac{1}{2014}\), \(\dfrac{1}{2017}< \dfrac{1}{2014}\)
\(\Rightarrow\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)< 0\\ \Rightarrow-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\\>0\\ \Rightarrow4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]>4\)
VN
1
TK
9 tháng 4 2016
A-3=2014/2015+2015/2016+2017/2015-3
=>A-3=-1/2015-1/2016+2/2015
=>A-3=1/2015-1/2016
Vì 1/2015>1/2016
=>1/2015-1/2016>0
=>A-3>0
=>A>3
TC
1
TH
13 tháng 2 2018
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
2 tháng 5 2016
2014+2015+2016/2015+2016+2017<2014/2015+2015/2016+2016/2017
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 ~