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\(M=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2012}+\sqrt{2013}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2013}-\sqrt{2012}\)
\(=\sqrt{2013}-1\)
\(S=\dfrac{1}{\sqrt{1.2012}}+\dfrac{1}{\sqrt{2.2011}}+...+\dfrac{1}{\sqrt{2012.1}}>\dfrac{1}{\dfrac{1+2012}{2}}+\dfrac{1}{\dfrac{2+2011}{2}}+...+\dfrac{1}{\dfrac{2012+1}{2}}=\dfrac{2012}{\dfrac{2013}{2}}=\dfrac{4024}{2013}\)
= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)
= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)
= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)
= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)
= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)
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 :
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)
\(x=\dfrac{\sqrt{\sqrt{5}-2}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+1\right)}}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(x=\dfrac{1+\sqrt{5}-2}{\sqrt{3-\sqrt{5}}}-\left(\sqrt{2}-1\right)=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{6-2\sqrt{5}}}-\left(\sqrt{2}-1\right)\)
\(x=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{\left(\sqrt{5}-1\right)^2}}-\sqrt{2}+1=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\sqrt{2}+1=1\)
\(\Rightarrow x^{2012}+2x^{2013}+3x^{2014}=1^{2012}+2.1^{2013}+3.1^{2014}=6\)
2 = 1.2 => \(\dfrac{1}{2}\) = \(\dfrac{1}{1.2}\) = 1 - \(\dfrac{1}{2}\)
TT \(\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)
.................
=> VT = 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{\sqrt{2012-x}+2012}{\sqrt{2012-x}+2013}\)
Đặt \(\sqrt{2012-x}+2012=y\)
=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{y}{y+1}\)
=> \(\dfrac{x}{x+1}\) = \(\dfrac{y}{y+1}\)
=> x = y
<=> x = \(\sqrt{2012-x}+2012\)
<=> 2012 - x + \(\sqrt{2012-x}\) = 0
<=> \(\sqrt{2012-x}=0\)
<=> x = 2012
Ta có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Thế vô bài toán ta được
\(A=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2012}}-\dfrac{1}{\sqrt{2013}}=1-\dfrac{1}{\sqrt{2013}}\)
Ta có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n.\left(n+1\right)}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Sau đó thế vô bài toán và làm tiếp như bác ctv là ta hoàn thành bài toán!