so sanh A=98/99
B=98*99+1/98*99
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Ta có: 99x99=99x(98+1) = 99x98+99
98x100= 98x(99+1) = 98x99+98
Ta có: 98x99=98x99 mà 99> 98 => 99x99>98x100
\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)
\(\Rightarrow C>D\)
Ta có: A = 98/99 < 1
B = 98*99+1/98*99 > 1
=> A < B
Ta có: A = 98/99 < 1
B = 98*99+1/98*99 > 1
=> A < B