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 ?

98   <1

99

98.99+1     Vì 98.99+1 >98.99 nên 98.99+1   >1

98.99                                             98.99

Suy  ra: 98     <    98.99+1  

            99            98.99

A= \(\frac{98}{99}\)< \(1\)

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

=> A<1<B => A<B

Ta có:

99.99= (98+1) . 99 = 98.99+99

98.100 = 98.( 99+1) = 98.99 + 98

Do  98.99+99 > 98.99 + 98 

=> 99.99 > 98.100

Không làm phép nhân hãy so sánh tích: 99 x 99 với tích 98 x 100

98 x 99 = (98 + 1) x 99 = 98 x 99 + 99

98 x 100 = 98 x (99 + 1) = 98 x 99 + 98

Vậy 99 x 99 > 98 x 100

bằng nhau vì 99-1 = 98 còn 99+ 1 = 100

`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!

\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)

\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)

\(\Rightarrow C>D\)

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

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

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

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\(A=\frac{98^{99}+1}{98^{89}+1}>\frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=B\)

Vậy A>B