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Xét hiệu:
\(C=1-\dfrac{7^{2011}+1}{7^{2013}+1}=\dfrac{7^{2011}\left(7^2-1\right)}{7^{2013}+1}=\dfrac{48.7^{2011}}{7^{2013}+1}\)
\(D=1-\dfrac{7^{2013}+1}{7^{2015}+1}=\dfrac{7^{2013}\left(7^2-1\right)}{7^{2015}+1}=\dfrac{48.7^{2013}}{7^{2015}+1}\)
Ta có:
\(\dfrac{C}{D}=\dfrac{48.7^{2011}}{7^{2013}+1}\cdot\dfrac{7^{2015}+1}{48.7^{2013}}=\dfrac{7^{2015}+1}{\left(7^{2013}+1\right)\cdot7^2}=\dfrac{7^{2015}+1}{7^{2015}+49}< 1\)
=> C<D =>A>B
Mình ko bít có đúng ko nên sai đừng trách mình nhé !
\(A=\frac{7^{2011}+1}{7^{2013}+1}\)
\(7^2.A=\frac{7^{2013}+49}{7^{2013}+1}=\frac{7^{2013}+1+48}{7^{2013}+1}=\)\(\frac{7^{2013}+1}{7^{2013}+1}+\frac{48}{7^{2013}+1}=1\frac{48}{7^{2013}+1}\)
\(B=\frac{7^{2013}+1}{7^{2015}+1}\)
\(7^2.B=\)\(=\frac{7^{2015}+49}{7^{2015}+1}=\)\(\frac{7^{2015}+1+48}{7^{2015}+1}=\)\(\frac{7^{2015}+1}{7^{2015}+1}+\frac{48}{7^{2015}+1}=1\frac{48}{7^{2015}+1}\)
\(Vì\) \(1\frac{48}{7^{2013}+1}>1\frac{48}{7^{2013}+1}\)\(\Rightarrow7^2.A>7^2.B\)\(\Rightarrow A>B\)
\(Vậy\) \(A>B\)
Bài 2 nè
ta xét B trước:
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..\)\(.....+\frac{1}{2015}-\frac{1}{2016}\)
=\(\left(\frac{1}{1}+\frac{1}{3}+....+\frac{1}{2015}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{2016}\right)\)
\(=\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}\right)-\)\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
vậy A:B\(=\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)\(:\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2016}\)
\(=1\)
a) Giải
Ta có: \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}+\dfrac{1}{2^{2013}}\)
\(\Rightarrow2S=\dfrac{2}{2}+\dfrac{2}{2^2}+\dfrac{2}{2^3}+...+\dfrac{2}{2^{2012}}+\dfrac{2}{2^{2013}}\)
\(2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2S-S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2012}}-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=1-\dfrac{1}{2^{2013}}\)
\(\Rightarrow S=\dfrac{2^{2013}-1}{2^{2013}}\)
b) Giải
Từ \(A=\dfrac{2011^{2012}+1}{2011^{2013}+1}\)
\(\Rightarrow2011A=\dfrac{2011^{2013}+20111}{2011^{2013}+1}=\dfrac{2011^{2013}+1+2010}{2011^{2013}+1}=1+\dfrac{2010}{2011^{2013}+1}\)
Từ \(B=\dfrac{2011^{2013}+1}{2011^{2014}+1}\)
\(\Rightarrow2011B=\dfrac{2011^{2014}+2011}{2011^{2014}+1}=\dfrac{2011^{2014}+1+2010}{2011^{2014}+1}=1+\dfrac{2010}{2011^{2014}+1}\)
Vì 20112013 + 1 < 20112014 + 1 và 2010 > 0
\(\Rightarrow\dfrac{2010}{2011^{2013}+1}>\dfrac{2010}{2011^{2014}+1}\)
\(\Rightarrow2011A>2011B\)
\(\Rightarrow A>B\)
Vậy A > B.
A= 1+2-3-4+5+6-7-8+...+2013+2014
A=(1+2-3-4)+(5+6-7-8)+.....+(2013+2014)
A=(-4)+(-4)+...+(-4)+4027
A=(-4).503+4027
A=-2012+4027
A=2015
B=\(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2016}{2015}\)
B=\(\dfrac{3.4.5.6.....2016}{2.3.4.5.....2015}=\dfrac{2016}{2}=1008\)
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
\(a,\frac{20132013}{20142014}=\frac{2013.10001}{2014.10001}=\frac{2013}{2014}=1-\frac{1}{2014};\frac{131313}{141414}=\frac{13.10101}{14.10101}=\frac{13}{14}=1-\frac{1}{14}.\text{Vì: 14 bé hơn 2014 nên:}\frac{1}{14}>\frac{1}{2014}\Rightarrow\frac{20132013}{20142014}>\frac{131313}{141414}\)
\(C=2013^9+2013^9.2013=2013^9\left(2013+1\right)=2013^9.2014;D=2014^9.2014\text{ vì: 2013^9< 2014^9 nên: C bé thua D }\)
\(c,M=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}};N=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2005}}.Vì:10^{2006}>10^{2005}.Nên:\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\Rightarrow M>N\)
b, Ta có:
\(14A=\dfrac{7^{2013}+14}{7^{2013}+1}=\dfrac{7^{2013}+1+13}{7^{2013}+1}=\dfrac{7^{2013}+1}{7^{2013}+1}+\dfrac{13}{7^{2013}+1}=1+\dfrac{13}{7^{2013}+1}\)
\(14B=\dfrac{7^{2015}+14}{7^{2015}+1}=\dfrac{7^{2015}+1+13}{7^{2015}+1}=\dfrac{7^{2015}+1}{7^{2015}+1}+\dfrac{13}{7^{2015}+1}=1+\dfrac{13}{7^{2015}+1}\)
\(\)Vì \(7^{2013}+1< 7^{2015}+1\)
\(\dfrac{\Rightarrow13}{7^{2013}+1}>\dfrac{13}{7^{2015}+1}\)
\(\Rightarrow1+\dfrac{13}{7^{2013}+1}>1+\dfrac{13}{7^{2015+1}}\)
\(\Leftrightarrow14A>14B\)
\(\Rightarrow A>B\)