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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 tháng 4 2017

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 \(A>B\)

\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)

=>x+2013=0

hay x=-2013

24 tháng 2 2022

\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)

\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)

AH
Akai Haruma
Giáo viên
11 tháng 12 2018

Lời giải:

Áp dụng BĐT Cô-si ngược dấu:

\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4(x-2010)}\leq \frac{4+(x-2010)}{4}\)

\(\Rightarrow \sqrt{x-2010}-1\leq \frac{4+(x-2010)}{4}-1=\frac{x-2010}{4}\)

\(\Rightarrow \frac{\sqrt{x-2010}-1}{x-2010}\leq \frac{1}{4}\)

Hoàn toàn tương tự với những phân thức còn lại:

\(\Rightarrow \frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}+\frac{\sqrt{z-2012}-1}{z-2012}\leq \frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-2010=4\\ y-2011=4\\ z-2012=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2014\\ y=2015\\ z=2016\end{matrix}\right.\)