Có \(\frac{2016}{2017}=1-\frac{1}{2017}\Rightarrow\frac{2016}{2017}+\frac{1}{2017}=1\)1
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
mà 1 = 1 và 2017 < 2018 nên \(\frac{1}{2017}>\frac{1}{2018}\)
suy ra \(\frac{2016}{2017}< \frac{2017}{2018}\)mặc khác \(\frac{2016}{2017}>\frac{1}{2017}\)nên\(\frac{2017}{2018}>\frac{1}{2017}\)do đó \(\frac{2016}{2017}+\frac{2017}{2018}>1\)
Cho A= \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2016}\)
So sanh A voi 3
a. So sanh 2 phan so:A= 2015/2016+2016/2017+2017/2018 va B = 2015+2016+2017/2016+2017+2018
b.1/2.4+1/4.6+........+1/(2x-2).2x = 1/8
c.Cho A = 1/4+1/9+1/16+...+1/81+1/100 . Chung minh rang : A > 65/132
d.Cho B = 12/(2 . 4 ) ^ 2 + 20/ (4 . 6) ^2 + ...........+ 388/ ( 96 . 98 ) ^ 2 + 396/ ( 98 . 100 ) ^2 .Hay so sanh B voi 1 /4
So sanh :\(\frac{2017}{2018}+\frac{2018}{2019}va\frac{2015}{2016}+\frac{2016}{2017}\)
so sanh 2016^2017 va 2017^2016
giup to voi!!!
Tinh A= 2014/2015+2015/2016+2016/2017+2017/2014 hay so sanh A voi 4
so sanh
A=\(\frac{2016^{2016}+1}{2016^{2017}+1}\)
B=\(\frac{2016^{2017}-3}{2016^{2018}-3}\)
2017/2016va 2018/2016 so sanh dau do la phan so
so sanh A voi 1 biet A= 2^2019-(2^2018+2^2017+...+2^1+2^0)
Cho tong T=2/2^1+3/2^2+4/2^3 +...+2016/2^2015+2017/2^2016.So sanh T voi 3
