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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
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 \(\frac{2012}{2013}< \frac{2011}{2012}\)
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
\(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
\(B=\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}<\frac{2011}{2012}+\frac{2012}{2013}=A\)
vậy A>B
\(A=\frac{2011}{2012}+\frac{2012}{2013}\) \(và\) \(B=\frac{2011+2012}{2012+2013}\)
\(Ta\) \(có\) \(:\) \(B=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}\)
\(B=\frac{2011}{4025}+\frac{2012}{4025}\)
\(Vì\) \(\frac{2011}{2012}>\frac{2011}{4025}và\frac{2012}{2013}>\frac{2012}{4025}\)
\(Nên\) \(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{4025}+\frac{2012}{4025}\)
\(Vậy\) \(A=\frac{2011}{2012}+\frac{2012}{2013}>B=\frac{2011+2012}{2012+2013}\)
a) \(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{3}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3}{12}+\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3+1+4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\dfrac{2}{3}-\dfrac{2011}{2012}=\dfrac{298}{719}\cdot\dfrac{3}{2}-\dfrac{2011}{2012}=\dfrac{149.3}{719.1}-\dfrac{2011}{2012}=\dfrac{447}{719}-\dfrac{2011}{2012}=\dfrac{889364}{1446628}-\dfrac{1445909}{1446628}=\dfrac{889364-1445909}{1446628}=-\dfrac{556545}{1446628}.\)b)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)
\(\dfrac{2011\cdot2012-1}{2011\cdot2012}=\dfrac{2011\cdot2012}{2011\cdot2012}-\dfrac{1}{2011\cdot2012}=1-\dfrac{1}{2011\cdot2012}\)
\(\dfrac{2012\cdot2013-1}{2012\cdot2013}=\dfrac{2012\cdot2013}{2012\cdot2013}-\dfrac{1}{2012\cdot2013}=1-\dfrac{1}{2012\cdot2013}\)
Vì \(\dfrac{1}{2011\cdot2012}>\dfrac{1}{2012\cdot2013}\Rightarrow1-\dfrac{1}{2011\cdot2012}>1-\dfrac{1}{2012\cdot2013}\)
Vậy \(\dfrac{2011\cdot2012-1}{2011\cdot2012}< \dfrac{2012\cdot2013-1}{2012\cdot2013}\)