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Xét dạng tổng quát: \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\sqrt{n}.\frac{1}{\sqrt{n}}+\sqrt{n}.\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2.\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Thay vào đề bài ta có:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2013\sqrt{2012}}\)
\(< 2.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(< 2.\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2013}}\right)\)
\(< 2.\left(1-\frac{1}{\sqrt{2013}}\right)< 2\left(đpcm\right)\)
......................?
mik ko biết
mong bn thông cảm
nha ................
Ta có :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
Ta có:
\(\frac{1}{\sqrt{x}+\sqrt{x-1}}=\frac{\sqrt{x}-\sqrt{x-1}}{\left(\sqrt{x}+\sqrt{x-1}\right)\left(\sqrt{x}-\sqrt{x-1}\right)}=\sqrt{x}-\sqrt{x-1}\)
Do đó:
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(\Leftrightarrow A=\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{n-1}+\sqrt{n}\)
\(\Leftrightarrow A=\sqrt{n}-1\left(dpcm\right)\)
ĐKXĐ: ...
a/ \(A=x-2009-4\sqrt{x-2009}+4=\left(\sqrt{x-2009}-2\right)^2\ge0\)
\(A_{min}=0\) khi \(\sqrt{x-2009}-2=0\Rightarrow x=2013\)
b/ \(\frac{1}{4}-\frac{\sqrt{x-2009}-1}{x-2009}+\frac{1}{4}-\frac{\sqrt{y-2010}-1}{y-2010}+\frac{1}{4}-\frac{\sqrt{z-2011}-1}{z-2011}=0\)
\(\Leftrightarrow\frac{x-2009-4\sqrt{x-2009}+4}{4\left(x-2009\right)}+\frac{y-2010-4\sqrt{y-2010}+4}{4\left(y-2010\right)}+\frac{z-2011-4\sqrt{z-2011}+4}{4\left(z-2011\right)}=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2009}-2\right)^2}{4\left(x-2009\right)}+\frac{\left(\sqrt{y-2010}-2\right)^2}{4\left(y-2010\right)}+\frac{\left(\sqrt{z-2011}-2\right)^2}{4\left(z-2011\right)}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)
Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
\(=\frac{\sqrt{n+1}}{\sqrt{n}.\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thay n = 1, 2, 3, ..., 2011 vào C ta có:
\(C=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
Vậy \(C=1-\frac{1}{\sqrt{2012}}.\)
uk xie xie (cảm ơn ) bạn , nhưng mik giải ra lâu r