CÁCH 1

Ta có \(A=\frac{89}{99}=\frac{99-1}{99}=\frac{99}{99}-\frac{1}{99}=1-\frac{1}{99}\)

\(B=\frac{98.99+1}{98.99}=\frac{98.99}{98.99}+\frac{1}{98.99}\)

Vì \(\frac{1}{98.99}< \frac{1}{99}\Rightarrow1+\frac{1}{98.99}>1-\frac{1}{99}\Rightarrow\frac{98.99+1}{98.99}>\frac{98}{99}\Rightarrow B>A\)

CÁCH 2 

Ta thấy 98 < 99 nên \(\frac{98}{99}< 1\)hay \(A< 1\)

Ta thấy \(98.99+1>98.99\Rightarrow\frac{98.99}{98.99+1}>1\Rightarrow B>1\)

Vì A < 1 ; B > 1 nên A < B

\(A=\frac{98}{99}< 1;\Rightarrow A< 1\)

\(B=\frac{98.99+1}{98.99}\)

Ta loại các số chia hết cho nhau thì được

\(B=\frac{1.1+1}{1.1}=1+1=2\)

\(2>1;\Rightarrow B>1;\Rightarrow B>A\)

\(A=\frac{98}{99}=1-\frac{1}{99}< 1\)

\(B=\frac{98.99+1}{99.98}=\frac{98.99}{99.98}+\frac{1}{99.98}=1+\frac{1}{99.98}>1\)

Vậy  \(A< B\)

p/s: chúc bạn học tốt

Ta có : \(\frac{98.99+1}{99.98}>\frac{98.99}{99.98}=1\)

\(\frac{98}{99}< 1\)

\(=>\frac{98.99+1}{99.98}>\frac{98}{99}\)

Ta có: B= \(\frac{98.99+1}{98.99}\)=\(\frac{99+1}{99}\)=\(\frac{100}{99}\)

Lại có: \(\frac{98}{99}\)< \(\frac{100}{99}\) nên \(\frac{98}{99}\)< \(\frac{98.99+1}{99.98}\)

Vậy suy ra A<B

*Mình không chắc là đúng đâu nha 

a,) a < b

b) a > b

c, a > b

Ko tính kết quả.Mình cam đoan luôn.

Chúc bạn học tốt `~<>

Monfan sub bạn cậu có thể trình bày ra cho tớ dc ko ?

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

Do \(\frac{98}{99}< 1\)

Mà \(\frac{99.98+1}{98.99}=\frac{9703}{9702}>1\)

Nên \(\frac{98}{99}< \frac{99.98+1}{98.99}\)

Vậy A < B

Ai thấy tớ đúng k nha

Ta có
\(A=\frac{98}{99}=\frac{98.98}{99.98};B=\frac{99.98+1}{98.99}\)
\(\Rightarrow98.98< 99.98+1\)
\(\Rightarrow\frac{98.98}{99.98}< \frac{99.98+1}{98.99}\)
\(\Rightarrow A< B\)

Ta có:

\(\frac{98}{99}< 1\)

\(\frac{99.98+1}{98.99}=\frac{1}{98.99}+1>1\)

Suy ra : A<B

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

`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!