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Lời giải:
\(A=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2011}{1.2.3...2012}\)
\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{2012-1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2011}-\frac{1}{1.2.3...2012}\)
\(=1-\frac{1}{1.2...2012}< 1\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :\(\left(2011+1\right)^2=2011^2+1+2.2011\)
\(\Rightarrow2011^2+1=2012-2.2011\)
\(\Rightarrow N=\sqrt{2012^2-2.2011+\left(\dfrac{2011}{2012}\right)^2}+\dfrac{2011}{2012}\)
\(=\sqrt{\left(2012-\dfrac{2011}{2012}\right)^2}+\dfrac{2011}{2012}\)
\(=2012-\dfrac{2011}{2012}+\dfrac{2011}{2012}\)
\(=2019\)
Vậy N có giá trị là một số tự nhiên.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2011\\y\ge2012\\z\ge2013\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-2011}\ge0\\b=\sqrt{y-2012}\ge0\\c=\sqrt{z-2013}\ge0\end{matrix}\right.\) ta có :
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\left\{\begin{matrix} x-2010=4\\ y-2011=4\\ z-2012=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2014\\ y=2015\\ z=2016\end{matrix}\right.\)