so sánh \(C=\frac{^{98^{99}}+1}{98^{89}+1}\)và D=98^98+1/98^88+1
khẩn cấp nhé
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LL
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)
\(\Rightarrow C>D\)
TD
0
NT
0
NT
33
G
28 tháng 2 2016
Lấy C - D
\(C-D=\frac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{98}+1\right)\left(98^{89}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
Tử số bằng:
\(98^{187}+98^{99}+98^{88}+1-98^{187}-98^{98}-98^{89}-1\)
=\(98^{99}+98^{88}-98^{98}-98^{89}\)
= \(98^{99}-98^{98}+98^{88}-98^{89}\)
= \(98^{98}\left(98-1\right)+98^{88}\left(1-98\right)\)
= \(98^{98}.97-98^{88}.97=97\left(98^{98}-98^{88}\right)>0\)
Vậy C - D > 0 => C > D
28 tháng 2 2016
Do C>1 nên ta có:
C=9899+1/9889+1>9899+1+97/9889+1+97=9899+98/9889+98=98(9898+1)/98(9888+1)=9898+1/9888+1=D
suy ra 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
DL
4 tháng 4 2018
\(C=\frac{98^{99}+1}{98^{89}+1}\)
\(D=\frac{98^{98}+1}{98^{88}+1}\)
\(C< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98^{98}\left(98+1\right)}{98^{88}\left(98+1\right)}\)
\(C< \frac{98^{98}}{98^{88}}=D\)
Ta có:C=\(\frac{98^{99}+1}{98^{89}+1}\Rightarrow\frac{98^{99}+1}{98^{99}+10}=\frac{98^{99}+1}{98^{99}+1+9}=\frac{98^{99}+1}{1+9}\)
D\(\frac{98^{98}+1}{98^{88}+1}=\frac{98^{98}+1}{98^{98}+10}=\frac{98^{98}+1}{98^{98}+1+9}\frac{98^{98}+1}{1+9}\)
Vì\(\frac{98^{99}+1}{1+9}\)>\(\frac{98^{98}+1}{1+9}\)
=>C>D