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19A=192010+19/192010+1=192010+1+18/192010+1=192010+1/192010+1+18/192010+1=1+18/192010
19B=192009+19/192009+1=192009+1+18/192009+1=192009+1/192009+1+18/192009+1=1+18/192009
Vậy A<B
Xin lỗi mình chịu câu trên
Ta có A=\(\frac{19^{2009}+1}{19^{2010}+1}\) Ta có:B=\(\frac{19^{2008}+1}{19^{2009}+1}\)
19B=\(\frac{19^{2009}+19}{19^{2009}+1}\)
19A=\(\frac{19^{2010}+19}{19^{2010}+1}\) 19B=\(\frac{19^{2009}+1+18}{19^{2009}+1}\)
19A=\(\frac{19^{2010}+1+18}{19^{2010}+1}\) 19B=\(1+\frac{18}{19^{2009}+1}\)
19A=\(1+\frac{18}{19^{2010}+1}\)
Vì \(\frac{18}{19^{2010}+1}< \frac{18}{19^{2009}+1}\)nên \(19A< 19B\)
\(\Leftrightarrow A< B\)
Vậy\(A< B\)
Đây nhá:
A=\(\frac{2009^{20}+9}{2009^{20}-1}=\frac{2009^{20}-1+10}{2009^{20}-1}=1+\frac{10}{2009^{20}-1}\)
B bạn tự làm như trên sẽ ra:
\(B=1+\frac{10}{2009^{19}-1}\)
Ta có: \(\frac{10}{2009^{20}-1}< \frac{10}{2009^{19}-1}\)
\(\Rightarrow\)A<B
1.
a) \(-\frac{15}{17}>-\frac{19}{21}\)
b)\(-\frac{13}{19}>-\frac{19}{23}\)
c)\(-\frac{23}{49}>-\frac{25}{47}\)
d)\(\frac{317}{633}>\frac{371}{743}\)
e)\(-\frac{24}{35}< -\frac{19}{30}\)
f)\(\frac{12}{17}< \frac{13}{18}\)
g) \(-\frac{17}{26}< -\frac{16}{27}\)
h) \(\frac{84}{-83}< -\frac{337}{331}\)
i) \(-\frac{1941}{1931}< -\frac{2011}{2001}\)
j) \(-\frac{1930}{1945}>-\frac{1996}{2001}\)
k) \(\frac{37}{59}< \frac{47}{59}\)
I) \(-\frac{25}{124}>-\frac{27}{100}\)
m) \(-\frac{97}{201}>-\frac{194}{309}\)
n) \(-\frac{189}{398}< -\frac{187}{394}\)
o) \(-\frac{289}{403}>-\frac{298}{401}\)
A=2016/2017+2017/2018
Do 2016/2017<1,2017/2018<1=> A<2 Hay A<B