NT
0
K
Khách
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TD
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
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
NT
1
1 tháng 3 2016
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
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
SG
3
17 tháng 3 2018
Vì C= \(\dfrac{98^{99}+1}{98^{89}+1}\)>1 thì nên áp dụng tính chất . Nên \(\dfrac{a}{b}\)>1 thì \(\dfrac{a}{b}\)>\(\dfrac{a+m}{b+m}\) ( a∈ N , b và m ∈ N✳) Ta có : C= \(\dfrac{98^{99}+1}{98^{89}+1}\)> \(\dfrac{98^{99}+1+97}{98^{89}+1+97}\)= \(\dfrac{98^{99}+98}{98^{89}+98}\) = \(\dfrac{98.98^{98}+98.1}{98.98^{88}+98.1}\) = \(\dfrac{98.\left(98^{98}+1\right)}{98.\left(98^{88}+1\right)}\)= \(\dfrac{98^{98}+1}{98^{88}+1}\)= B ⇔ Vậy \(\dfrac{98^{99}+1}{98^{89}+1}\)< \(\dfrac{98^{89}+1}{98^{88}+1}\) nên C<D