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\(\frac{199}{200}>\frac{199}{200+201+202}\)
\(\frac{200}{201}>\frac{200}{200+201+202}\)
\(\frac{201}{202}>\frac{201}{200+201+202}\)
=>\(A>B\)
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
200+201/201+202=200/403+201/403
vì 200/201>200/403
201/202>201/403 nên \(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)
mik làm câu A thôi nha
ta có :
1 - 2009/2010 = 1/2010
1 - 2010/2011 = 1/2011
Phần bù nào bé thì phân số đó lớn .
Vì 1/2010 > 1/2011
Nên 2009/2010 > 2010/2011
Ta thấy hiệu giữa mẫu số và tử số của hai phân số bằng nhau ( = 1 )
Để so sánh hai phân số, ta so sánh các hiệu.
\(1-\frac{2009}{2010}\)và \(1-\frac{2010}{2011}\)
Ta có :
\(1-\frac{2009}{2010}=\frac{2010}{2010}-\frac{2009}{2010}=\frac{1}{2010}\)
\(1-\frac{2010}{2011}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
Ta thấy :
\(\frac{1}{2010}>\frac{1}{2011}\)
Hay :
\(1-\frac{2009}{2010}>1-\frac{2010}{2011}\)
Vậy \(\frac{2009}{2010}< \frac{2010}{2011}\)
\(\frac{200+201}{201+202}=\frac{200}{201+202}+\frac{201}{201+201}\)
Mà \(201\frac{200}{201+202}\)
\(\frac{201}{202}>\frac{201}{201+202}\)
=> \(\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)
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
\(\frac{200}{201}+\frac{201}{202}=1,99...>1>\frac{401}{403}=\frac{200+201}{201+202}\)
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
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 \(\frac{200}{201}>\frac{200}{201+202},\frac{201}{202}>\frac{201}{201+202}\Rightarrow\frac{200}{201}+\frac{201}{202}>\frac{200+201}{201+202}\)