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BV
0
L
1
10 tháng 3 2021
Ta có \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018};\frac{2016}{2017}>\frac{2016}{2016+2017+2018};\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\) nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Hay \(A>B\)
TA
1
DM
10 tháng 4 2016
ta có 2015/2016+2016/2017+2017/2015=(1-1/2016)+(1-1/2017)+(2+1/2015)
=4-(1/2016+1/2017-1/2015)
1/2016<1; 1/2017<1 nên 1/2016+1/2017<2 suy ra 1/2016+1/2017-1/2015<1(vì 1/2015<1)
4-(1/2016+1/2017-1/2015)>4-1=3
2015/2016+2016/2017+2017/2015>3
cho mik nhé
13 tháng 8 2017
a, Ta có :
\(A=\dfrac{15}{14}+\dfrac{16}{15}+\dfrac{17}{16}+\dfrac{18}{17}\)
\(\Leftrightarrow A=\left(1+\dfrac{1}{14}\right)+\left(1+\dfrac{1}{15}\right)+\left(1+\dfrac{1}{16}\right)+\left(1+\dfrac{1}{17}\right)\)
\(\Leftrightarrow A=\left(1+1+1+1\right)+\left(\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}\right)\)
\(\Leftrightarrow A=4+\left(\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}\right)\)
\(\Leftrightarrow A>4\)
b. \(B=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2019}\)
\(\Leftrightarrow B=\left(1-\dfrac{1}{2016}\right)+\left(1-\dfrac{1}{2017}\right)+\left(1-\dfrac{3}{2019}\right)\)
\(\Leftrightarrow B=\left(1+1+1\right)-\left(\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{3}{2019}\right)\)
\(\Leftrightarrow B=3-\left(\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{3}{2019}\right)\)
\(\Leftrightarrow B< 3\)
HT
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 \(\dfrac{1}{2015}< \dfrac{1}{2014}\), \(\dfrac{1}{2016}< \dfrac{1}{2014}\), \(\dfrac{1}{2017}< \dfrac{1}{2014}\)
\(\Rightarrow\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)< 0\\ \Rightarrow-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\\>0\\ \Rightarrow4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]>4\)
\(S=\frac{1015}{2016}+\frac{2016}{2017}+\frac{2021}{2018}=\frac{1016-1}{2016}+\frac{2017-1}{2017}+\frac{2018+3}{2018}\)
=> \(S=1-\frac{1}{2016}+1-\frac{1}{2017}+1+\frac{3}{2018}=3+\left(\frac{3}{2018}-\frac{1}{2016}-\frac{1}{2017}\right)\)
Nhận thấy; \(\frac{3}{2018}-\frac{1}{2016}-\frac{1}{2017}>0\)=> S > 3