So sánh C=\(\frac{98^{99}+1}{98^{89}+1}\) và D=\(\frac{98^{98}+1}{98^{88}+1}\)
So sánh C=\(\frac{98^{99}+1}{98^{89}+1}\) và D=\(\frac{98^{98}+1}{98^{88}+1}\)
So sánh C=\(\frac{98^{99}+1}{98^{89}+1}\) và D=\(\frac{98^{98}+1}{98^{88}+1}\)
so sánh
\(\frac{98^{99}+1}{98^{89}+1}\)và \(\frac{-98^{98}-1}{-98^{88}-1}\)
\(A=\frac{-\left(98^{98}+1\right)}{-\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}\)
\(B=\frac{98^{99}+1}{98^{89}+1}\)
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
So sánh :
\(C=\frac{98^{99}+1}{98^{89}+1}\) và \(D=\frac{98^{98}+1}{98^{88}+1}\)
Ai làm nhanh và đúng mk sẽ tick cho !!!
So sánh :
C= \(\dfrac{98^{99}+1}{98^{89}+1}\) và D = \(\dfrac{98^{98}+1}{98^{88}+1}\)
\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)
\(\Rightarrow C>D\)
so sánh \(C=\frac{^{98^{99}}+1}{98^{89}+1}\)và D=98^98+1/98^88+1
khẩn cấp nhé
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
So sánh
C= \(\frac{98^{99}+1}{98^{89}+1}\)
D=\(\frac{98^{98}+1}{98^{88}+1}\)
Giải theo cách tính \(\frac{1}{98^{10}}C\)và \(\frac{1}{98^{10}}D\)
1) So sánh : A= \(\frac{17^{18}+1}{17^{19}+1}\) và B = \(\frac{17^{17}+1}{17^{18}+1}\)
2) So sánh: C = \(\frac{98^{99}+1}{98^{89}+1}\)và D = \(\frac{98^{98}+1}{98^{88}+1}\)
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
So sánh 2 phân số sau:
\(A=\frac{98^{99}+1}{98^{89}+1}\) và\(B=\frac{98^{98}+1}{98^{88}+1}\)
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