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2010/2011+2011/2012+2012/2013+671/670 > 4

17 tháng 5 2022

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

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

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

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

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

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

\(P>Q\)

10 tháng 5 2017

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

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

5 tháng 7 2023

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

\(\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)

Vì \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\) 

=> \(x+2014=0\) 

                  \(x=0-2014\) 

                  \(x=-2014\)

21 tháng 3 2023

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)

`=> (x+2014) (1/2010 + 1/2011-1/2012-1/2013)=0`

`=> x+2014=0` ( vì `1/2010 + 1/2011-1/2012-1/2013≠0 )`

`=>x=-2014`