AR
2
K
Khách
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NK
2
21 tháng 5 2015
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Rightarrow A>B\)
KH
13 tháng 3 2017
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Theo đề ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Lúc này ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Leftrightarrow A>B\).
MT
9 tháng 7 2016
@@@) Ta có: \(A=\frac{5^{2016}+4}{5^{2015}+4}\Rightarrow\frac{1}{5}A=\frac{5^{2016}+4}{5^{2016}+20}=1+\frac{-16}{5^{2016}+20}\)
\(B=\frac{5^{2014}+4}{5^{2013}+4}\Rightarrow\frac{1}{5}B=\frac{5^{2014}+4}{5^{2014}+20}=1+\frac{-16}{5^{2014}+20}\)
Ta thấy: \(1+\frac{-16}{5^{2016}+20}>1+\frac{-16}{5^{2014}+20}\) =>\(\frac{1}{5}A>\frac{1}{5}B\Rightarrow A>B\)
Bài thứ 2 sai để nhé hai cái đó = nhau mà
NT
0
N
15 tháng 7 2018
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+17}{17^{19}+17}\)
\(A=\frac{17^{18}+1}{17^{19}+1}< \frac{17^{17}+1}{17^{18}+1}=B\)
=> A < B
S
11 tháng 3 2017
Bài 1:
Ta thấy A < 1
=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)
Vậy A < B
Bài 2:
Ta thấy C < 1
=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C < D
HP
1
TN
27 tháng 4 2017
\(A=\frac{-\left(98^{98}+1\right)}{-\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}\)
\(B=\frac{98^{99}+1}{98^{89}+1}\)
A-1=\(\frac{98^{98}-98^{88}}{98^{88}+1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}\)
B-1=\(\frac{98^{99}-98^{89}}{98^{89}+1}=\frac{98^{89}.\left(98^{10}-1\right)}{98^{89}+1}\)
=>\(\frac{A-1}{B-1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}.\frac{98^{89}+1}{98^{89}.\left(98^{10}-1\right)}=\frac{98^{89}+1}{98.\left(98^{88}+1\right)}=\frac{98^{89}+1}{98^{89}+98}< 1\)
->A-1<B-1
->A<B
13 tháng 4 2016
A=\(\frac{98^{99}+1}{98^{89}+1}>1\) =>\(A=\frac{98^{99}+1}{98^{89}+1}>\frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}\)
\(=\frac{98.\left(98^{98}+1\right)}{98.\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C>D
Có \(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
nên \(A=\frac{98^{2015}+1}{98^{2014}+1}>\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014+98}}\)\(=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}=B\)
Vậy A>B