Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014

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Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)

\(\sqrt{y-2014}=b\left(b>0\right)\)

\(\sqrt{z-2015}=c\left(c>0\right)\)

Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)

<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)

<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)

<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).

Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)

\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)

\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)

=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)

Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)

ko bt

\(\frac{1}{\sqrt{2013}-\sqrt{2014}}-\frac{1}{\sqrt{2014}-\sqrt{2015}}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2014}-\sqrt{2015}\right)}-\frac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)

\(=\sqrt{2015}-\sqrt{2013}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)

\(=\sqrt{2015}-\sqrt{2013}\)

Em mới lớp 6 thui! Sorry vì ko giúp đc

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lục sách tìm !có đấy ! tuy em có mới học lớp 5 nhưng thấy qua bai này rùi !!!!!

\(\left(\sqrt{2014}-\sqrt{2013}\right).\left(\sqrt{2014}+\sqrt{2013}\right)\)

=> \(\sqrt{2014^2}-\sqrt{2013^2}\)

=> \(2014-2013\)

\(=1\)

Vậy ..............

\(\left(\sqrt{2014}-\sqrt{2013}\right).\left(\sqrt{2014}+\sqrt{2013}\right)=1\)

\(VT=\left(\sqrt{2014}-\sqrt{2013}\right).\left(\sqrt{2014}+\sqrt{2013}\right)\)

\(=\sqrt{2014^2}-\sqrt{2013^2}\)

\(=2014-2013\)

\(=1=VP\left(dpcm\right)\)

Vậy....

\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}+...-\frac{1}{\sqrt{2013}-\sqrt{2014}}+\frac{1}{\sqrt{2014}-\sqrt{2015}}\)

\(=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{\sqrt{3}+\sqrt{4}}{3-4}+...+\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{3}+\sqrt{4}-\left(\sqrt{4}+\sqrt{5}\right)+...+\sqrt{2014}+\sqrt{2015}\)

=\(-\sqrt{2}+\sqrt{2015}\)

* Cách 1: 

\(\sqrt{1^2+2013^2+\frac{2013^2}{2014^2}}\)

\(=\sqrt{2013^2.\left(1+\frac{1}{2013^2}+\frac{1}{2014^2}\right)}\)

\(=2013.\left(1+\frac{1}{2013}-\frac{1}{2014}\right)\)

\(=2013+1-\frac{2013}{2014}\)

\(=2014-\frac{2013}{2014}\)

* Cách 2:

\(\sqrt{1^2+2013^2+\frac{2013^2}{2014^2}}\)

\(=\sqrt{\left(1+2013\right)^2-2.2013+\frac{2013^2}{2014^2}}\)

\(=\sqrt{2014^2-2.2013+\left(\frac{2013}{2014}\right)^2}\)

\(=\sqrt{\left(2014-\frac{2013}{2014}\right)^2}\)

\(=2014-\frac{2013}{2014}\)

Từ 2 cách trên ta suy ra:

\(\sqrt{1^2+2013^2+\frac{2013^2}{2014^2}}+\frac{2013}{2014}\)

\(=2014-\frac{2013}{2014}+\frac{2013}{2014}\)

\(=2014\)

Theo đề bài trên, ta có thể suy ra công thức tổng quát như sau:

\(\sqrt{1^2+x^2+\frac{x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)

(Chúc bạn học tốt và nhớ k cho mình với nhá!)

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