Câu hỏi của Nguyễn Tuấn Minh - Toán lớp 6 - Học toán với OnlineMath

\(A=1-\frac{1}{2013.2014}\) ; \(B=1-\frac{1}{2014.2015}\)

Vì \(\frac{1}{2013.2014}>\frac{1}{2014.2015}\) nên A < B

\(\frac{2013\cdot2014-1}{2013\cdot2014}=\frac{2013\cdot2013}{2013\cdot2014}=\frac{2013}{2014}\)

\(\frac{2014\cdot2015-1}{2014\cdot2015}=\frac{2014\cdot2014}{2014\cdot2015}=\frac{2014}{2015}\)

Vậy \(\frac{2014}{2015}>\frac{2013}{2014}\)

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Ta có:A=(2013.2014-1)/2013.2014

            =1-(1/2013.2014)

Ta lại có:B=(2014.2015-1)/2014.2015

                 =1-(1/2014.2015)

Dễ thấy:1/2013.2014>1/2014.2015

             =>1-(1/2013.2014)<1-(1/2014.2015)

A=-2015/2015x2016

A=-1/2016

B=-2014/2014x2015

B=-1/2015

vi 2016>2015,-1/2016>-1/2015

vay A>B

b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)

\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)

Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)

\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)

Ta có: \(10^{2010}+1< 10^{2011}+1\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)

\(\Leftrightarrow10A>10B\)

hay A>B

a) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)\)

\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-99}{100}\) có 99 số hạng

\(=-\frac{1}{100}\)

b) \(A=\frac{2013.2014-1}{2013.2014}=1-\frac{1}{2013.2014}\)

\(B=\frac{2014.2015-1}{2014.2015}=1-\frac{1}{2014.2015}\)

Vì 2013.2014 < 2014.2015 

=> \(\frac{1}{2013.2014}>\frac{1}{2014.2015}\)

=> \(1-\frac{1}{2013.2014}< 1-\frac{1}{2014.2015}\)

=> A < B 

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)

\(=1-\dfrac{1}{2015}\)

\(=\dfrac{2014}{2015}\)

Vì \(2014.2015=2014.2015\)nên \(2014.2015-1< 2014.2015\)1 đơn vi

Vì \(2015.2016=2015.2016\)nên \(2015.2016-1< 2015.2016\)1 đơn vị

Ta có :

\(1-M=1-\frac{2014.2015-1}{2014.2015}=\frac{1}{2014.2015}\)

\(1-N=1-\frac{2015.2016-1}{2015.2016}=\frac{1}{2015.2016}\)

Vì \(2015=2015\)nên \(2014.2015< 2015.2016\)

Vì \(\frac{1}{2014.2015}>\frac{1}{2015.2016}\)( do \(2014.2015< 2015.2016\))

Nên \(N>M\)

Vậy \(N>M\)