tính A biết A=(\(\dfrac{\sqrt[4]{2010^2}-\sqrt[4]{2010}}{1-\sqrt[4]{2010}}+\dfrac{1+\sqrt{2010}}{\sqrt[4]{2010}}\)) -\(\dfrac{\sqrt{1+\dfrac{2}{2010}+\dfrac{1}{2010}}}{1+\sqrt{2010}}\)
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S
30 tháng 7 2019
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+........+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\frac{1}{\sqrt{1}\sqrt{2}\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{\sqrt{2}\sqrt{3}\left(\sqrt{2}+\sqrt{3}\right)}+........+\frac{1}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2009}+\sqrt{2010}\right)}\)
\(=\frac{\left(\sqrt{2010}-\sqrt{2009}\right)\left(\sqrt{2010}+\sqrt{2009}\right)}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2010}+\sqrt{2009}\right)}+.......+\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}=1-\frac{1}{\sqrt{2010}}=1-\frac{\sqrt{2010}}{2010}\)
PA
26 tháng 2 2018
\(\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{\sqrt{z-2011}-1}{z-2011}=\dfrac{3}{4}\)\(\left(\left\{{}\begin{matrix}x>2009\\y>2010\\z>2011\end{matrix}\right.\right)\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{1}{4}-\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{1}{4}-\dfrac{\sqrt{z-2011}-1}{z-2011}=0\)
\(\Leftrightarrow\dfrac{x-2009-4\sqrt{x-2009}+4}{x-2009}+\dfrac{y-2010-4\sqrt{y-2010}+4}{y-2010}+\dfrac{z-2011-4\sqrt{z-2011}+4}{z-2011}=0\)
Nhận xét: \(\left\{{}\begin{matrix}\dfrac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}\ge0\\\dfrac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}\ge0\\\dfrac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(2013;2014;2015\right)\)
1 tháng 3 2018
\(\Leftrightarrow\dfrac{4\sqrt{x-2009}-4}{x-2009}-1+\dfrac{4\sqrt{x-2009}-4}{x-2009}-1+\dfrac{4\sqrt{x-2009}-4}{x-2009}-1=0\)\(\Leftrightarrow-\dfrac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}-\dfrac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}-\dfrac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}=0\)
VT <=0 đẳng thức khi và chỉ khi \(\left\{{}\begin{matrix}x-2009=4=>x=2013\\y=2014\\z=2015\end{matrix}\right.\)
1 tháng 5 2018
Đặt a = \(\sqrt{x-2009}\)
b = \(\sqrt{y-2010}\)
c = \(\sqrt{z-2011}\)
\(\Leftrightarrow\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{a}-\dfrac{1}{a^2}+\dfrac{1}{b}-\dfrac{1}{b^2}+\dfrac{1}{c}-\dfrac{1}{c^2}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{a}-\dfrac{1}{a^2}-\dfrac{1}{4}+\dfrac{1}{b}-\dfrac{1}{b^2}-\dfrac{1}{4}+\dfrac{1}{c}-\dfrac{1}{c^2}-\dfrac{1}{4}=0\)
\(\Leftrightarrow-(\dfrac{1}{a}-\dfrac{1}{2})^2-\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2-\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)
Dấu = xảy ra khi
a = 2
b = 2
c = 2
\(\Leftrightarrow\sqrt{x-2009}=2\)
\(\sqrt{y-2010}=2\)
\(\sqrt{z-2011}=2\)
\(\Leftrightarrow x-2009=4\)
\(y-2010=4\)
\(z-2011=4\)
=> x = 2013
y = 2014
z = 2015
31 tháng 1 2019
Lời giải:
Ta có $$\frac{\sqrt{x-2009}-1}{x-2009}+\frac{\sqrt{y-2010}-1}{y-2010}+\frac{\sqrt{z-2011}-1}{z-2011}=\frac{3}{4} \Leftrightarrow \left ( \frac{1}{\sqrt{x-2009}}-\frac{1}{2} \right )^2+\left ( \frac{1}{\sqrt{y-2010}}-\frac{1}{2} \right )^2+\left ( \frac{1}{\sqrt{z-2011}}-\frac{1}{2} \right )^2=0$$
$$\Rightarrow x=2013,y=2014,z=2015$$ 
TD
11 tháng 11 2017
Mình chỉ viết CT tổng quát thôi nha rồi bạn tự thay vào
a, \(\frac{1}{\sqrt{n}(n+1)+n\sqrt{n+1} }=\frac{1}{\sqrt{n(n+1)( }\sqrt{n}+\sqrt{n+1}} =\frac{\sqrt{n+1}-\sqrt{n} }{\sqrt{n}\sqrt{n+1} } =\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } \)
b,\(\frac{1}{\sqrt{n}+\sqrt{n+1} }=\frac{\sqrt{n+1}-\sqrt{n} }{1}= \sqrt{n+1}-\sqrt{n} \)