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nghịch dảo 2 p/s ta có:
\(\frac{2010}{2009}v\text{à}\frac{2011}{2010}\)
\(\frac{2010}{2009}=1\frac{1}{2009}\) ;\(\frac{2011}{2010}=1\frac{1}{2010}\)
vì \(\frac{1}{2009}>\frac{1}{2010}\)
=>\(\frac{2009}{2010}
1-\(\frac{2009}{2010}=\frac{2010}{2010}-\frac{2009}{2010}=\frac{1}{2010}\)
1-\(\frac{2010}{2011}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
Vì 2010 < 2011 nên \(\frac{1}{2010}>\frac{1}{2011}nên\frac{2009}{2010}>\frac{2010}{2011}\)
Vậy \(\frac{2009}{2010}>\frac{2010}{2011}\)
\(\dfrac{2011}{2010}>1;\dfrac{2010}{2011}< 1\\ Nên:\dfrac{2011}{2010}>1>\dfrac{2010}{2011}\\ Vậy:\dfrac{2011}{2010}>\dfrac{2010}{2011}\)
Mình chỉ làm được câu a thôi!!!!
2010/2009 và 2011/2010
Ta có:
2010-1+1/2009=2009+1/2009=2009/2009+1/2009=1+1/2009
2011-1+1/2010=2010+1/2010=2010/2010+1/2010=1+1/2010
Vì 2009<2010=>1/2009>1/2010
=>1+1/2009>1+1/2010
=>2010/2009>2011/2010
Vậy 2010/2009>2011/2010
\(\dfrac{2011}{2010}=1-\dfrac{1}{2010}\)
\(\dfrac{97}{96}=1-\dfrac{1}{96}\)
mà -1/2010>-1/96
nên 2011/2010>97/96
\(\dfrac{2011}{2010}=1-\dfrac{1}{2010}\)
\(\dfrac{97}{96}=1-\dfrac{1}{96}\)
mà 1/2010<1/96
nên 2011/2010>97/96
\(A=\frac{m^{2010}+1}{m^{2011}+1};B=\frac{m^{2011}+1}{m^{2012}+1}\)
Ta có:
\(A=\frac{m^{2010}+1}{m^{2011}+1}\Rightarrow10A=\frac{m^{2011}+10}{m^{2011}+1}\)
\(B=\frac{m^{2011}+1}{m^{2012}+1}\Rightarrow10B=\frac{m^{2012}+10}{m^{2012}+1}\)
Hay ta so sánh: \(\frac{9}{m^{2011}};\frac{9}{m^{2012}}\)
Vì \(2011< 2012\)nên \(m^{2011}< m^{2012}\)hay \(\frac{9}{m^{2011}}>\frac{9}{m^{2012}}\)
Vậy \(A>B\)
A = 1-1/2011+1-1/2012 = 2-(1/2011+1/2012) > 1 ( vì 1/2011+1/2012 < 1 )
B = 4021/4023 = 1-2/4023 < 1
=> A > B
k mk nha
Ta có:\(\frac{192}{2011}< \frac{192}{2010}\)vì 2011>2010 (1)
\(\frac{192}{2010}< \frac{193}{2010}\)vì 192<193 (2)
Từ (1) và (2) ta có: \(\frac{192}{2011}< \frac{193}{2010}\)
Quy đồng 2 mâu số ta có :
\(\frac{192}{2011}=\frac{192.2010}{2011.2010}=\frac{385920}{4042110}\)
\(\frac{193}{2010}=\frac{193.2011}{2010.2011}=\frac{388123}{4042110}\)
Vậy \(\frac{385920}{4042110}< \frac{388123}{4042110}\left(385920< 388123\right)\) . Nên \(\frac{192}{2011}< \frac{193}{2010}\)