Đặt A=1998/1999+1999/2000 B=1998+1999/1999+2000 =1998/1999+2000 + 1999/1999+2000 Vì 1998/1998>1998/1999+2000 1999/2000>1999/1999+2000 Nên A>B

Đặt A=1998/1999+1999/2000 
B=1998+1999/1999+2000
=1998/1999+2000 + 1999/1999+2000
Vì 1998/1998>1998/1999+2000
1999/2000>1999/1999+2000
Nên A>B

Mình chịu

a) Có vẻ đề o đúng lắm . Theo mình o phải là 11/11 mà 1/11

Ta có \(\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>...>\frac{1}{19}>\frac{1}{20}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

hay \(S>\frac{1}{2}\)

b)Ta có 1998 x 1999 + 3997=(2000-2) x 1999 +3997 = 2000 x 1999 - 2 x 1999 +3997 = 1999 x 2000 -3998 +3997 =1999 x 2000 -1

< 1999 x 2000 +2 

=> 1999 x 2000 +2 / 1998 x 1999 +3997 > 1 hay M>1

Thanks you . Mình sẽ kết bạn với cậu nhé

a)1999/2001<1

12/11>1

=>1999/2001<12/11

b)

1998/1999=1-1/1999

1999/2000=1-1/2000

Vì 1/1999>1/2000

=>1998/1999<1999/2000

A) 1 - 1974/1975 = 1/1975

1 - 1998/1999 = 1/ 1999

Mặt khác 1/1975 > 1/1999

=> 1974/1975 < 1998/1999

Ta có 1974/1975 =1-1974/1975=1975/1975-1

974/1975=1/1979.

1998/1999 =1-1998/1999=1999/1999-1998/1999 =1/1999. 

Vì 1/1979 l/h 1/1999 nên 1974 /1975 l/h 1998/1999.

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So sánh

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )

Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)

Vậy B > A

Chúc bạn học tốt

\(\left(\sqrt{1998}+\sqrt{2000}\right)^2=1998+2000+2.1998.2000=2.1999+2.1998.2000\)

\(\left(2\sqrt{1999}\right)^2=4.1999=2.1999+2.1999\)

Mà  \(2.1998.2000>2.1999\)

\(=>\left(\sqrt{1998}+\sqrt{2000}\right)^2>\left(2\sqrt{1999}\right)^2=>\sqrt{1998}\)+\(\sqrt{2000}>2\sqrt{1999}\)

Câu trả lời đây bạn nhé ^^

http://olm.vn/hoi-dap/question/602523.html

Ta có:

\(A=1998\times1998=1998^2\)

\(B=1996\times2000=\left(1998-2\right)\left(1998+2\right)\)

\(=1998^2-2^2=1998^2-4\)

Vì \(1998^2>1998^2-4\Rightarrow A>B\)

bạn có thể làm cho mk  câu 1 đc ko 

xin bạn đấy

1999x2000-2/1998x1999+3997

=1999x(1998+2)-2/1998x1999+3997

=1999x1998+3998-2/1998x1999+3997

=1999x1998+3996/1998x1999+3997

=3996/3997

=> 3996/3997 < 1

Vậy 1999x2000-2/1998x1999+3997 < 1

3998 ở dâu vậy