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\(Q=\dfrac{2010+2011+2012}{2011+2012+2013}=\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
Ta có: \(\dfrac{2010}{2011+2012+2013}< \dfrac{2010}{2011}\)
\(\dfrac{2011}{2011+2012+2013}< \dfrac{2011}{2012}\)
\(\dfrac{2012}{2011< 2012< 2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(P>Q\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\cdot x=\left(1+\dfrac{2011}{2}\right)+\left(1+\dfrac{2010}{3}\right)+...+\left(\dfrac{1}{2012}+1\right)+1\)
\(\Leftrightarrow x\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2013}\)
=>x=2013
Đặt \(B=A\div C\)
\(C=2012+\dfrac{2011}{2}+...+\dfrac{1}{2012}=2012+\dfrac{2013-2}{2}+\dfrac{2013-3}{3}+...+\dfrac{2013-2012}{2012}\)
\(C=2012+\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2012}-1-1-...-1\)
\(C=2012+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)-2011\)
\(C=1+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)=\dfrac{2013}{2013}+2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)\)
\(C=2013\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=2013.A\)
\(\Rightarrow B=\dfrac{A}{C}=\dfrac{1}{2013}\)
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.
Bài 1:
1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà \(17^{19}+1>17^{18}+1\)
nên 17A>17B
hay A>B
2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)
\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)
mà \(98^{89}+1>98^{88}+1\)
nên C>D
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
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
DỄ THẤY A<1
B=(2011.2013+2012.2012)/2012.2013
ta có 2011.2013+2012.2012-2012.2013=2012.2012+2013.(2011-2012)
=2012.2012-2013
suy ra 2011.2013+2012.2012>2012.2013
=> B >1 mà A <1
SUY RA B>A
B = 2011/2012+2012/2013 > 2011/2013+ 2012/2013 = 2011+2012/2013>2011+2012/ 2012+2013= A.
Vậy B>A
Bài 1:
Ta có: \(A=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)
Dễ thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow A=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< B=\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
Bài 2:
\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)