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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.
\(A=\dfrac{2011^{2011}+2}{2011^{2011}-1}=\dfrac{2011^{2011}-1+3}{2011^{2011}-1}=\dfrac{2011^{2011}-1}{2011^{2011}-1}+\dfrac{3}{2011^{2011}-1}=1+\dfrac{3}{2011^{2011}-1}\left(1\right)\)
\(B=\dfrac{2011^{2011}}{2011^{2011}-3}=\dfrac{2011^{2011}-3+3}{2011^{2011}-3}=\dfrac{2011^{2011}}{2011^{2011}}+\dfrac{3}{2011^{2011}-3}=1+\dfrac{3}{2011^{2011}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
1+\(\dfrac{3}{2011^{2011}-1}\)>1+\(\dfrac{3}{2011^{2011}-3}\)mà
1) \(\dfrac{1}{2011}+\dfrac{2012.2010}{2011}-2012\)=\(\dfrac{1+2012.2010-2012.2011}{2011}\)
= \(\dfrac{1+2012.\left(2010-2011\right)}{2011}\)= \(\dfrac{1+2012.\left(-1\right)}{2011}\)
= \(\dfrac{-2011}{2011}=-1\)
2.A=\(\dfrac{43.11}{2011^{2013}}\)+\(\dfrac{79}{2011^{2013}}\)=\(\dfrac{43.11+79}{2011^{2013}}\)
B=\(\dfrac{79.11}{2011^{2013}}\)+\(\dfrac{43}{2011^{2013}}\)=\(\dfrac{79.11+43}{2011^{2013}}\)
Ta có: 43.11+79=43.(10+1)+79=43.10+43+79=430+122
79.11+43=79.(10+1)+43=79.10+79+43=790+122
Vì 430+122<790+122 nên 43.11+79<79.11+43 (1)
Mà 20112013<20112013 (2)
Từ (1) và (2) suy ra A<B
3. A=\(\dfrac{2010.2012}{2011.2011}\)
Vì B<1 nên B>\(\dfrac{2010}{2012}\)=\(\dfrac{2010.2012}{2012.2012}\)
Vì 2010.2012=2010.2012; 2011.2011<2012.2012 nên B>A
4. A=\(\dfrac{3n}{3\left(2n+1\right)}\)=\(\dfrac{3n}{6n+3}\)
Vì 6n+3=6n+3; 3n<3n+1 nên A<B
a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
\(Q=\dfrac{1}{2011}+\dfrac{2}{2010}+\dfrac{3}{2009}+...+\dfrac{2010}{2}+\dfrac{2011}{1}\)
\(Q=\left(1+\dfrac{2}{2011}\right)\left(1+\dfrac{2}{2010}\right)+\left(1+\dfrac{3}{2009}\right)+...+\left(1+\dfrac{2010}{2}\right)+1\)
\(Q=\dfrac{2012}{2011}+\dfrac{2012}{2010}+\dfrac{2012}{2009}+...+\dfrac{2012}{2}+\dfrac{2012}{2012}\)
\(Q=2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(\Rightarrow\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)}=\dfrac{1}{2012}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{1.3}\)
\(...\)
\(\dfrac{1}{100^2}>\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{100}=\dfrac{99}{100}\\ \dfrac{99}{100}< \dfrac{1}{4}\\ \Rightarrowđpcm\)
a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.