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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: \(\dfrac{2010}{2011+2012+2013}< \dfrac{2010}{2011}\)
\(\dfrac{2011}{2011+2012+2013}< \dfrac{2011}{2012}\)
\(\dfrac{2012}{2011< 2012< 2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)
\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(P>Q\)
\(Q=\dfrac{2010+2011+2012}{2011+2012+2013}=\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)Ta thấy:
\(\dfrac{2010}{2011}>\dfrac{2010}{2011+2012+2013}\\ \dfrac{2011}{2012}>\dfrac{2011}{2011+2012+2013}\\ \dfrac{2012}{2013}>\dfrac{2012}{2011+2012+2013}\\ \Rightarrow\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\\ \Leftrightarrow P>Q\)
Vậy \(P>Q\)
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:
\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)
Ta có : \(B=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}=\dfrac{2012}{2012+2013}\)
Mà : \(\dfrac{2011}{2012}>\dfrac{2011}{2012+2013}\)
\(\dfrac{2012}{2013}>\dfrac{2012}{2012+2013}\)
\(\Rightarrow \dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)
\(\Rightarrow\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011+2012}{2012+2013}\)
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