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\(\dfrac{2011\cdot2012-1}{2011\cdot2012}=\dfrac{2011\cdot2012}{2011\cdot2012}-\dfrac{1}{2011\cdot2012}=1-\dfrac{1}{2011\cdot2012}\)
\(\dfrac{2012\cdot2013-1}{2012\cdot2013}=\dfrac{2012\cdot2013}{2012\cdot2013}-\dfrac{1}{2012\cdot2013}=1-\dfrac{1}{2012\cdot2013}\)
Vì \(\dfrac{1}{2011\cdot2012}>\dfrac{1}{2012\cdot2013}\Rightarrow1-\dfrac{1}{2011\cdot2012}>1-\dfrac{1}{2012\cdot2013}\)
Vậy \(\dfrac{2011\cdot2012-1}{2011\cdot2012}< \dfrac{2012\cdot2013-1}{2012\cdot2013}\)
Ta có \(\frac{2012.2013}{2012.2013+1}\)và \(\frac{2013}{2012}\)
Vì \(\frac{2012.2013}{2012.2013+1}< 1< \frac{2013}{2012}\)
nên \(\frac{2012.2013}{2012.2013+1}< \frac{2013}{2012}\)
\(\frac{2012}{2013}\)và \(\frac{2011}{2012}\)
phàn bù của \(\frac{2012}{2013}\)là \(\frac{1}{2013}\)
phàn bù của \(\frac{2011}{2012}\)là \(\frac{1}{2012}\)
Vì \(\frac{1}{2012}>\frac{1}{2013}\Rightarrow\frac{2012}{2013}>\frac{2011}{2012}\)
Ta có : \(\frac{2012\cdot2013}{2012\cdot2013+1}< 1\)
\(\frac{2013}{2012}>1\)
\(\Rightarrow\frac{2012\cdot2013}{2012\cdot2013+1}< \frac{2013}{2012}\)
Có : \(\frac{2012}{2013}=1-\frac{2012}{2013}=\frac{2013}{2013}-\frac{2012}{2013}=\frac{1}{2013}\)
\(\frac{2011}{2012}=1-\frac{2011}{2012}=\frac{2012}{2012}-\frac{2011}{2012}=\frac{1}{2012}\)
Vì \(2013< 2012\)nên \(\frac{1}{2013}< \frac{1}{2012}\)hay \(\frac{2012}{2013}< \frac{2011}{2012}\)
a, \(A=\frac{2012\cdot2011-1}{2010\cdot2012+2011}=\frac{2012\cdot\left(2010+1\right)-1}{2010\cdot2012+\left(2012-1\right)}=\frac{2012\cdot2010+2012-1}{2012\cdot2010+2012-1}=1\)
b, 10,11 + 11,12 + 12,13 + .... + 97,98 + 98,99 + 99,100
= ( 10 + 11 + 12 + .... + 97 + 98 + 99 ) + ( 0,10 + 0,11 + 0,12 + 0,13 + ... + 0,98 + 0,99 )
= { ( 10 + 99 ) . [ ( 99 - 10 ) : 1 + 1 ] ] : 2 } + { ( 0,10 + 0,99 ) . [ ( 0,99 - 0,10 ) : 0,01 + 1 ] : 2 }
= ( 99 . 90 : 2 ) + ( 1,09 . 90 : 2 )
= 4455 + 49,05
= 4504,05
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(=\left(1+\frac{1}{2}+...+\frac{1}{2014}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\)
\(B=\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
\(=\frac{1}{3022}\left(\frac{3022}{1008.2014}+\frac{3022}{1009.2013}+...+\frac{3022}{2014.1008}\right)\)
\(=\frac{1}{3022}\left(\frac{1008}{1008.2014}+\frac{2014}{1008.2014}+...+\frac{2014}{1008.2014}+\frac{1008}{1008.2014}\right)\)
\(=\frac{1}{3022}\left(\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+...+\frac{1}{2014}+\frac{1}{1008}\right)\)
\(=\frac{2}{3022}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)\)
\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)\)
=> \(\frac{A}{B}=\frac{\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2014}\right)}=\frac{1}{\frac{1}{1511}}=1511\)
Vậy....
\(\frac{2010}{2011}\)> \(\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}\)> \(\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}\)> \(\frac{2012}{2011+2012+2013}\)
=> \(\frac{2010}{2011}\)+ \(\frac{2011}{2012}\)+ \(\frac{2012}{2013}\)> \(\frac{2010+2011+2012}{2011+2012+2013}\)
=> P > Q
Ta có :
\(Q=\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Vì :
\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\)\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010+2011+2012}{2011+2012+2013}\)
\(\Rightarrow\)\(P>Q\)
Vậy \(P>Q\)
Chúc bạn học tốt ~
Bài làm
\(A=\frac{2011.2012-1}{2011.2012}\) và \(B=\frac{2012.2013-1}{2012.2013}\)
Ta có:
\(A=\frac{2011.2012-1}{2011.2012}\)
\(A=\frac{2011.2012-1.1-1.1}{2011.2012}\)
\(A=\frac{2011.2012-1.\left(1-1\right)}{2011.2012}\)
\(A=\frac{2011.2012-1.0}{2011.2012}\)
\(A=\frac{2011.2012-0}{2011.2012}\)
\(A=\frac{2011.2012}{2011.2012}\)
\(A=1\)
\(B=\frac{2012.2013-1}{2012.2013}\)
\(B=\frac{2012.2013-1.1-1.1}{2012.2013}\)
\(B=\frac{2012.2013-1.\left(1-1\right)}{2012.2013}\)
\(B=\frac{2012.2013-1.0}{2012.2013}\)
\(B=\frac{2012.2013-0}{2012.2013}\)
\(B=\frac{2012.2013}{2012.2013}\)
\(B=1\)
Vì 1 = 1
=> A = B
Hay
\(A=\frac{2011.2012-1}{2011.2012}\)= \(B=\frac{2012.2013-1}{2012.2013}\)
Vậy \(A=\frac{2011.2012-1}{2011.2012}\)= \(B=\frac{2012.2013-1}{2012.2013}\)
# Chúc bạn học tốt #
Ta có : A =( 2011.2012-1)/(2011.2012) = (2011.2012)/(2011.2012) - 1/(2011.2012) = 1 - (1/2011.2012)
B =( 2012.2013-1)/(2012.2013) = (2012.2013)/(2012.2013) - 1/(2012.2013) = 1 - (1/2012.2013)
Ta thấy : 1/(2011.2012)>1/(2012.2013)(vì chung tử số là 1 , mẫu số : 2011.2012 < 2012.2013)
Suy ra , 1-(1/2011.2012)<1-(1/2012.2013)
Suy tiếp : A < B