`A=3/4+8/9+.............+9999/10000`

`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`

`=99-(1/4+1/9+.........+1/10000)<99-0=99`

`=>A<99`

 địt mẹ con ngu t khinh

Giải:

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\) 

\(A=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{8}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{10000}\right)\) 

\(A=99-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{19}+...+\dfrac{1}{10000}\right)< 99\) 

\(\Rightarrow A< 99\left(đpcm\right)\) 

Chúc bạn học tốt!

đởn giản  chúng đều= nhau

\(\frac{19^{100}+1}{19^{99}+1}< \frac{19^{100}+1+18}{19^{99}+1+18}=\frac{19.\left(19^{99}+1\right)}{19.\left(19^{98}+1\right)}=\frac{19^{99}+1}{19^{98}+1}\)

\(\Rightarrow A< B\)

Vậy A<B

k minh nha

Tính chất nếu: 

\(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) 

Ta có:

\(A=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}\)

\(A>\dfrac{10^{99}+10}{10^{89}+10}\)

\(A>\dfrac{10\cdot\left(10^{98}+1\right)}{10\cdot\left(10^{88}+1\right)}\)

\(A>\dfrac{10^{98}+1}{10^{88}+1}\)

\(A>B\)

\(A=\dfrac{10^{99}+1}{10^{89}+1}< \dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{89}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}\)

Vậy \(A< B\)

\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)

\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\) 

\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)

Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\) 

1) 19/19 < 2005/2004

2)72/73 > 98/99

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