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

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

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

b)2014/2014*2015=2014:2014/2014*2015:2014=1/2015(rút gọn phân số)

    2015/2015*2015=2015:2015/2015*2016:2015=1/2016(rút gọn phân số)

Mà 1/2015>1/2016

=>2014/2014*2015>2015/2015*2015

b.2014/2014×2015 và 2015/2015×2016

Ta có:

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

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

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

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

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

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

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

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

\(hnhaminhhlai\)

a)

Vì \(\frac{2009}{2010}< 1\Rightarrow\frac{2009}{2010}< \frac{2009+1}{2010+1}=\frac{2010}{2011}\)

Cần nhớ:

Nếu: \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\left(n\inℕ^∗\right)\)

Và tương tự:  \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\left(n\inℕ^∗\right)\)

b)Ta có:

 \(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)

\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)

Vì: \(81^{100}>64^{100}\Leftrightarrow\frac{1}{81^{100}}< \frac{1}{64^{100}}\Leftrightarrow\frac{1}{3^{400}}< \frac{1}{4^{300}}\)

c) Ta có:

\(\frac{200+201}{201+202}=\frac{401}{403}< 1\)

\(\frac{200}{201}+\frac{201}{202}=1-\frac{1}{201}+1-\frac{1}{202}=2-\left(\frac{1}{201}+\frac{1}{202}\right)>1\)

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

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

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

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

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

Dễ thấy 200+201/201+202<1

Ta có 200/201=1-1/201;201/202=1-1/202

=>200/201+201/202=1-1/201+1-1/202

=1-1/201+1-1/202

=(1+1)-(1*201+1/202)

=2-(1/201+1/202)

Đễ thấy 1/201+1/202<1

=>2-(1/201+1/202)>1

Mà 200+201/201+202<1

=>200/201+201/202>200+201/201+202

Câu này của bạn Hà Như Thuỷ bị nhầm mà.

Phải là: \(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+202}\)

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

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

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

=>\(A>B\)

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