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Ta có: \(\frac{2011\times2012-2}{2010\times2011+4020}=\frac{2011\times\left(2010+2\right)-2}{2010\times2011+4020}=\frac{2011\times2010+4022-2}{2010\times2011+4020}=\frac{2011\times2010+4020}{2010\times2011+4020}=1\)
Vậy \(\frac{2011\times2012-2}{2010\times2011+4020}=1\)
\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2010.2011+2.2011-2}{2010.2011+4020}=\frac{2010.2011+4042-2}{2010.2011+4020}=\frac{2010.2011+4040}{2010.2011+4040}=1\)
Vậy P = 1
\(M=\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010}{2011+2012}+\frac{2011}{2011+2012}=\frac{2010+2011}{2011+2012}=N\)
chắc chắn là A > B
hãy ủng hộ mk bằng một niềm tin nhé
^ _ ^ hihi
#)Giải :
Ta có : \(1-\frac{2010}{2011}=\frac{1}{2011}\)
\(1-\frac{2011}{2012}=\frac{1}{2012}\)
Vì \(\frac{1}{2011}>\frac{1}{2012}\Rightarrow\frac{2010}{2011}>\frac{2011}{2012}\)
\(\frac{2010}{2011}=1-\frac{1}{2011}\)
\(\frac{2011}{2012}=1-\frac{1}{2012}\)
\(2011\)<\(2012\)\(\Rightarrow\frac{1}{2011}\)>\(\frac{1}{2012}\)
\(\Rightarrow\frac{2010}{2011}\)<\(\frac{2011}{2012}\)
ta có: \(N=\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)
\(\Rightarrow\frac{2010}{2011}>\frac{2010}{2011+2012};\frac{2011}{2012}>\frac{2011}{2011+2012}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)
=> M>N
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:\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.2010+4022-2}{2010.2011+4020}=\frac{2010.2011+4020}{2010.2011+4020}=1\)