Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
chắc chắn là A > B
hãy ủng hộ mk bằng một niềm tin nhé
^ _ ^ hihi
\(A=\frac{2011\times2012}{2011+2012}+\frac{2009\times2010}{2009+2010}\)
\(A=\frac{2011\times2011}{2011+2012}+\frac{2011}{2011+2012}+\frac{2010\times2010}{2009+2010}-\frac{2010}{2009+2010}\)
\(A=\frac{2011\times2011}{2011+2012}+\frac{2010\times2010}{2009+2010}+\frac{2011}{2011+2012}-\frac{2010}{2009+2010}\)
\(A=B+\frac{2011}{2011+2012}-\frac{2010}{2009+2010}\)
\(A=B+\frac{2011}{4023}-\frac{2010}{4019}\)
Dễ thấy \(\frac{2011}{4023}-\frac{2010}{4019}< 0\)
\(\Rightarrow A< B\)
\(M=\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010}{2011+2012}+\frac{2011}{2011+2012}=\frac{2010+2011}{2011+2012}=N\)
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) 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:\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.2010+4022-2}{2010.2011+4020}=\frac{2010.2011+4020}{2010.2011+4020}=1\)
Ta có \(A=\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}\)
=> \(A=\left(1-\frac{1}{2010}\right)+\left(1-\frac{1}{2011}\right)+\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)\)
\(\Rightarrow A=\left(1+1+1+1\right)-\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)< 4=B\)
=> A < B