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chắc chắn là A > B
hãy ủng hộ mk bằng một niềm tin nhé
^ _ ^ hihi
Quy đồng
A=(2006.246813579+2007.987654321)/(246...
B=(2007.246813579+2006.987654321)/(246...
MS bằng nhau nên ta so sánh tử:
A'=2006.246813579+2007.987654321=2006.... +987654321
=2006.(246813579+987654321)+987654321
B'=2007.246813579+2006.987654321 =2006.246813579+246813579+2006.987654321
=2006.(246813579+987654321)+246813579
987654321>246813579
Nên A'>B'
Hay A>B
2007/2008<1
2008/2009<1
2009/2010<1
2010<2011<1
=>2007/2008+2008/2009+2009/2010+2010/2011<1+1+1+1
=>2007/2008+2008/2009+2009/2010+2010/2011<4(điều cần chứng minh)
2007/2008 < 1
2008/2009 < 1
2009/2010 < 1
2010/2011 < 1
=> 2007/2008 + 2008/2009 + 2009/2010 + 2010/2011 < 1 + 1 + 1 + 1
=>2007/2008 + 2008/2009 + 2009/2010 + 2010/2011 < 4 ( điều cần chứng minh )
ai tk mình mình tk lại cho
\(A=\frac{2011\times2012}{2011+2012}+\frac{2009\times2010}{2009+2010}\)
\(A=\frac{2011\times2011}{2011+2012}+\frac{2011}{2011+2012}+\frac{2010\times2010}{2009+2010}-\frac{2010}{2009+2010}\)
\(A=\frac{2011\times2011}{2011+2012}+\frac{2010\times2010}{2009+2010}+\frac{2011}{2011+2012}-\frac{2010}{2009+2010}\)
\(A=B+\frac{2011}{2011+2012}-\frac{2010}{2009+2010}\)
\(A=B+\frac{2011}{4023}-\frac{2010}{4019}\)
Dễ thấy \(\frac{2011}{4023}-\frac{2010}{4019}< 0\)
\(\Rightarrow A< B\)
a) Ta có : \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)=> M > N
b) P = \(\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.\left(2010+2\right)-2}{2010.2011+4020}=\frac{2011.2010+2011.2-2}{2010.2011+4020}=\)\(\frac{2011.2010+4020}{2010.2011+4020}=1\)
Nên P = 1
câu b sửa lại:\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.2010+4022-2}{2010.2011+4020}=\frac{2010.2011+4020}{2010.2011+4020}=1\)
\(\frac{2010}{2011}< \frac{2011}{2012}\)
\(\frac{11}{12}=\frac{22}{24}\)
\(\frac{25}{30}>\frac{25}{49}\)
\(\frac{1}{5}< \frac{3}{8}\)
\(\frac{1995}{1997}< \frac{1995}{1996}\)
Ta có : \(\frac{2009}{987654321}< \frac{2010}{987654321}\)
\(\frac{2010}{24681357}>\frac{2009}{24681357}\)
\(\Rightarrow A=B\)
\(A=\frac{2009}{987654321}+\frac{2010}{246813579}\)
\(=\frac{2009}{987654321}+\frac{2009}{246813579}+\frac{1}{246813579}\)
\(B=\frac{2009}{987654321}+\frac{1}{987654321}+\frac{2009}{246813579}\)
Có \(\frac{1}{246813579}>\frac{1}{987654321}\)
Vậy A > B