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Câu hỏi của Nguyễn Tuấn Minh - Toán lớp 6 - Học toán với OnlineMath
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 \(N>M\)
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
1. Ta có: \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(m\in Z\right)\)
\(B=\dfrac{2016^{2016}}{2016^{2016}-3}>\dfrac{2016^{2016}+2}{2016^{2016}-3+2}=\dfrac{2016^{2016}+2}{2016^{2016}-1}=A\)
Vậy A > B
2. \(\dfrac{1}{2016.2015}+\dfrac{1}{2015.2014}+\dfrac{1}{2014.2013}+...+\dfrac{1}{1.2}\)
= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}+\dfrac{1}{2015.2016}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)
= \(1-\dfrac{1}{2016}\)
=\(\dfrac{2015}{2016}\)
Ta có : \(A\text{=}\dfrac{2013.2014-1}{2013.2014}\text{=}\dfrac{2013.2014}{2013.2014}-\dfrac{1}{2013.2014}\text{=}1-\dfrac{1}{2013.2014}\)
\(B\text{=}\dfrac{2014.2015-1}{2014.2015}\text{=}\dfrac{2014.2015}{2014.2015}-\dfrac{1}{2014.2015}\text{=}1-\dfrac{1}{2014.2015}\)
\(Ta\) có : \(\dfrac{1}{2013.2014}>\dfrac{1}{2014.2015}\)
\(\Rightarrow A< B\)