\(Q=\dfrac{2010+2011+2012}{2011+2012+2013}=\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)

Ta có: \(\dfrac{2010}{2011+2012+2013}< \dfrac{2010}{2011}\)

           \(\dfrac{2011}{2011+2012+2013}< \dfrac{2011}{2012}\)

           \(\dfrac{2012}{2011< 2012< 2013}< \dfrac{2012}{2013}\)

\(\Rightarrow\dfrac{2010}{2011+2012+2013}+\dfrac{2011}{2011+2012+2013}+\dfrac{2012}{2011+2012+2013}\)

\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2013}\)

\(P>Q\)

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

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.

Ta có: \(A=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)

Vì \(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012}\) và \(\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)

\(\Rightarrow\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)

\(\Rightarrow\dfrac{2011+2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)

\(\Rightarrow A< B\)

Vậy \(A< B\).

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

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:

\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)

b)

\(B=\dfrac{2011^{2012}+1}{2011^{2012}-4}>\dfrac{2011^{2012}+1+3}{2011^{2012}-4+3}=\dfrac{2011^{2012}+4}{2011^{2012}-1}=A\)

Vậy B>A

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 \(A>B\)