Ta có:\(\frac{200}{201}>\frac{200}{201+202}và\frac{201}{202}>\frac{201}{201+202}\)

Suy ra\(\frac{200}{201}+\frac{201}{202}>\frac{200}{201+202}+\frac{201}{201+202}=\frac{200+201}{201+202}\)

Vậy\(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)

Ta co:\(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+202}\)

Vi \(\frac{200}{201}>\frac{200}{201+202},\frac{201}{202}>\frac{201}{201+202}\Rightarrow\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)

Ta có : \(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+201}\)

Mà : \(201<201+202\Rightarrow\frac{200}{201}>\frac{200}{201+202}\)

\(\frac{201}{202}>\frac{201}{201+202}\)

\(\Rightarrow\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)

\(hnhaminhhlai\)

\(\frac{199}{200}>\frac{199}{200+201+202}\)

\(\frac{200}{201}>\frac{200}{200+201+202}\)

\(\frac{201}{202}>\frac{201}{200+201+202}\)

=>\(A>B\)

Do \(\frac{199}{200}\)> \(\frac{199}{200+201+202}\), \(\frac{200}{201}\)>\(\frac{200}{200+201+202}\),\(\frac{201}{202}\)>\(\frac{201}{200+201+202}\)nên A>B

\(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}< \frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}\)

A                                                 \(< \frac{199+200+201}{200+201+202}=B\)

\(A< B\)

Ta có: \(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}< \frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}< \)

                                                              \(< \frac{199+200+201}{200+201+202}\)

Vậy A < B

ỦNG HỘ TỚ NHA

\(B=\frac{199+200+201}{200+201+202}\)

\(B=\frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}< \frac{199}{200}+\frac{200}{201}+\frac{201}{202}=A\)

Vậy B < A ( bài này chủ yếu so sánh hai phân số cùng từ, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn )

200+201 hay 200+202

2009/2010=1-1/2010<1-1/2011=2010/2011

vậy 2009/2010<2010/2011

3^400=(3^4)^100=81^100>64^100=4^300

=>1/3^400<1/4^300

Vậy 1/3^400<1/4^300

\(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+201}\)

Mà \(201\frac{200}{201+202}\)

\(\frac{201}{202}>\frac{201}{201+202}\)

=> \(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)

\(\frac{200}{201}+\frac{201}{202}=1,99...>1>\frac{401}{403}=\frac{200+201}{201+202}\)

\(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+201}\)

Mà \(201< 201+202\Rightarrow\frac{200}{201}>\frac{200}{201+202}\)

\(\frac{201}{202}>\frac{201}{201+202}\)

Vậy \(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)

Gọi d là UCLN(n,n+1)

Ta có:n+1 chia hết cho d

         n chia hết cho d

=>(n+1)-n chia hết cho d

=>1 chia hết cho d

=>d=1

Vậy phân số n/n+1 tối giản

ta co:(n,n+1)=dn

talai co:(n+1)-n=1 chia het cho d suy ra d=1.vayn/n+1 toi gian