2/5 x 1/X + 1/X x 2 = 0,1

1/X x ( 2/5 + 2 ) = 0,1

1/X x 12 / 5 = 0,1

1/X             = 0,1 :12/5 = 1/10 : 12/5

1/X             = 1/24

Vậy X = 24

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

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

199 - 200 + 201 - 202 + 203 - 204 + 205 - 206 + 207

= ( 199 - 200 ) + ( 201 - 202 ) + ( 203 - 204 ) + ( 205 - 206 ) + 207

= ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + 207

= -4 + 207

= 203

199 - 200 + 201 - 202 + 203 - 204 + 205 - 206 + 207 = 203 

199-200+201-202+203-204+205-206+207

=203

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