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TP
1
11 tháng 4 2019
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\left(1-\frac{1}{2016}\right)+\left(1-\frac{1}{2017}\right)+\left(1-\frac{1}{2018}\right)+\left(1+\frac{3}{2015}\right)\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(=\left(1+1+1+1\right)+\left(\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\right)\)
\(=4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)\)
Vì \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2015}>\frac{1}{2017};\frac{1}{2015}>\frac{1}{2018}\)
\(\Rightarrow\frac{1}{2015}-\frac{1}{2016}>0;\frac{1}{2015}-\frac{1}{2017}>0;\frac{1}{2015}-\frac{1}{2018}>0\)
\(\Rightarrow\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>0\)
\(\Rightarrow4+\left(\frac{1}{2015}-\frac{1}{2016}\right)+\left(\frac{1}{2015}-\frac{1}{2017}\right)+\left(\frac{1}{2015}-\frac{1}{2018}\right)>4\)
\(\Rightarrow A>4\left(dpcm\right)\)
LG
0
24 tháng 4 2019
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(A=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)\)
Xét :
\(\frac{1}{2016}< \frac{1}{2015}\)\(;\)\(\frac{1}{2017}< \frac{1}{2015}\)\(;\)\(\frac{1}{2018}< \frac{1}{2015}\)
\(\Rightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}< \frac{1}{2015}+\frac{1}{2015}+\frac{1}{2015}\)
\(\Leftrightarrow\)\(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}< 0\)
Suy ra : \(A=4-\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{3}{2015}\right)>4-0=4\) ( đpcm )
...
TT
1
YY
19 tháng 4 2018
\(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2015}\)
\(=\frac{2016-1}{2016}+\frac{2017-1}{2017}+\frac{2018-1}{2018}+\frac{2015+3}{2015}\)
\(=1-\frac{1}{2016}+1-\frac{1}{2017}+1-\frac{1}{2018}+1+\frac{3}{2015}\)
\(=4+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2015}-\frac{1}{2018}\)
mà \(\frac{1}{2015}>\frac{1}{2016};\frac{1}{2017};\frac{1}{2018}\)
\(\Rightarrow A>4\)
NT
0
PH
2
9 tháng 1 2020
A = 4 + 42 + 43 + ... + 42013
A = (4 + 42) + (43 + 44) + ... + (42012 + 42013)
A = (4 + 42) + 42(4 + 42) + ... + 42011(4 + 42)
A = 20 +(42.20)+ ... +(42011.20)
A = 20(1 + 42 + ... + 42011) ⋮ 5
➤ A ⋮ 5
A = 4 + 42 + 43 + ... + 42013
A =(4 + 42 + 43) + (44 + 45 + 46 )+ ...+ (42011 + 42012 + 42013)
A= (4 + 42 + 43) +43(4 + 42 + 43) + ... +42010(4 + 42 + 43)
A = 81 + (43.81) + ... +(42010.81)
A = 81(1 + 43 + ... + 42010) ⋮ 21
➤ A ⋮ 21
IN
26 tháng 2 2020
A=1+32+34+36+....+3100
=>9A=32+34+36+38+....+3102
=>8A=3102-1
=>A=3102-1/8
b)A=1+53+56+59+.....+599
125A=53+56+59+512+.....+5102
124A=5102-1
A=5102-1/124
BT3:
1+4+42+43+...+458+459
=>(1+4)+(42+43)+...........+(458+459)chia hết cho 5
=>5+42.5+...........+458.5 chia hết cho 5
2)1+4+42+43+........+458+459
=>(1+4+42)+(43+44+45)+..........+(457+458+459)
=>21+43.21+........+457.24 chia hết cho 21
3)1+4+42+43+..........+458+459
=>(1+4+42+43)+(44+45+46+47)+............+(456+457+458+459)
=>85+44.85+..........+456.85 chia hết cho 85
4)5+53+55+.........+5202+5203 ( đề sai vì ta thấy 53 tới 55 mà 5202 tới 5203)
A=4^2015+4^2016+4^2017+4^2018
A=(4^2015+4^2016)+(4^2017+4^2018)
A=4^2015.(1+4)+4^2017.(1+4)
A=5.(4^2015+4^2017)
=>A chia hết cho 5