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Ta có:
\(A=\dfrac{2010}{2011}+\dfrac{2011}{2012}\)
\(B=\dfrac{2010+2011}{2011+2012}\)
\(=\dfrac{2010}{2011+2012}+\dfrac{2011}{2011+2012}\)
Áp dụng tính chất \(\dfrac{a}{b}>\dfrac{a}{b+m}\) ta có:
\(\left\{{}\begin{matrix}\dfrac{2010}{2011}>\dfrac{2010}{2011+2012}\\\dfrac{2011}{2012}>\dfrac{2011}{2011+2012}\end{matrix}\right.\)
\(\Rightarrow\dfrac{2010}{2011}+\dfrac{2011}{2012}>\dfrac{2010}{2011+2012}+\dfrac{2011}{2011+2012}\)
Hay \(\dfrac{2010}{2011}+\dfrac{2011}{2012}>\dfrac{2010+2011}{2011+2012}\)
Vậy \(A>B\)
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:
\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)
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.
Mai Quỳnh
B = 2011/2012+2012/2013 > 2011/2013+ 2012/2013
= 2011+2012/2013>2011+2012/ 2012+2013
= A.
Vậy B>A
Ta có:\(A=\dfrac{2011+1012}{2012+2013}\)
\(A=\dfrac{2011}{4023}+\dfrac{2012}{4023}< \dfrac{2011}{2012}+\dfrac{2012}{2013}=B\)
=> A<B
Vậy A<B
Ta có
A=\(\dfrac{2011+2012}{2012+2013}\)=\(\dfrac{2011}{2012+2013}\)+\(\dfrac{2012}{2012+2013}\)(1)
B=\(\dfrac{2011}{2012}\)+\(\dfrac{2012}{2013}\)(2)
=>A>B
A lớn
B nhỏ
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.
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.
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