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Ta có:
2010 . 2011/2010 . 2011 + 1 2009 . 2010/2009 . 2010 + 1
= 1 - 1/2010 . 2011 + 1 = 1 - 1/2009 . 2010 + 1
Vì 2010 . 2011 + 1 > 2009 . 2010 + 1
=> 1/2010 . 2011 + 1 < 1/2009 . 2010 + 1
=> 1 - 1/2010 . 2011 + 1 > 1 - 1/2009 . 2010 + 1
=> 2010.2011/2010.2011+1 > 2009.2010/2009.2010+1
Ta có: 2009.2010>2008.2009
\(\frac{1}{2009\cdot2010}< \frac{1}{2008\cdot2009}\)
\(\Rightarrow E>F\)
2008.2009+4018
=2008.2009+2009.2
=2009.(2008+2)
=2009.3000
=6027000
b.2010.2011+4020
=2010.2011+2010.2
=2010.(2011+2)
=2010.2013
=4046130
2008 x 2009 + 40 18
= 4034072 + 4018
= 4038090
2010 x 2011 + 4020
= 4042110 + 4020
= 4046130
( 2008 + 1009 ) . ( 2009 - 1009 ) + 4018
3017 . 1000 + 4018
3017000 + 4018
3021018
A = \(\frac{2009.2010-2}{2008+2008.2010}=\frac{2009.2010-2}{2008.\left(2010+1\right)}=\frac{2009.2010-2}{2008.2011}=\frac{2008.2010+2010-2}{2008.2011}=\frac{2008.2011}{2008.2011}=1\)
B = \(\frac{-2009.20102010}{20092009.2010}=\frac{-2009.10001.2010}{2009.10001.2010}=-1\)
1 > -1 => A > B
Ta có:
\(A=\frac{2009.2010-2}{2008+2008.2010}\)
\(A=\frac{\left(2008+1\right).2010-2}{2008+2008.2010}\)
\(A=\frac{2008.2010+2010-2}{2008+2008.2010}\)
\(A=\frac{2008.2010+2008}{2008+2008.2010}\)
\(A=1\)
\(B=\frac{-2009.20102010}{20092009.2010}\)
\(B=\frac{-2009.2010.10001}{2009.10001.2010}\)
\(B=-1\)
Vì \(1>-1\Rightarrow A>B\)
Vậy \(A>B\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+....+100}\)
Các bạn làm ơn giải giúp mình với
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{100.101}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{100.101}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=2.\frac{99}{202}\)
\(=\frac{99}{101}\)
\(P=\frac{2009.2010-1}{2009.2010}=\frac{2009.2010}{2009.2010}-\frac{1}{2009.2010}=1-\frac{1}{2009.2010}\)
\(Q=\frac{2010.2011-1}{2010.2011}=\frac{2010.2011}{2010.2011}-\frac{1}{2010.2011}=1-\frac{1}{2010.2011}\)
Vì \(\frac{1}{2009.2010}>\frac{1}{2010.2011}\)nên \(1-\frac{1}{2009.2010}< 1-\frac{1}{2010.2011}\)\(\Leftrightarrow\)\(P< Q\)
p<q nha