<=>  1+\(\frac{1}{2014}\)+\(\frac{1}{x}\)=\(\frac{1}{x+1}\)+1+\(\frac{1}{2013}\)

<=>   \(\frac{1}{2014}\)+\(\frac{1}{x}\)=\(\frac{1}{x+1}\)+\(\frac{1}{2013}\)

<=>   \(\frac{1}{x}\)-\(\frac{1}{x+1}\)=\(\frac{1}{2013}\)-\(\frac{1}{2013+1}\)                  => x=2013

\(\frac{2015}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}\)

\(\Leftrightarrow\frac{2015}{2014}-1+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}-1\)

\(\Leftrightarrow\frac{1}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{1}{2013}\)

\(\Leftrightarrow\frac{x+2014}{2014x}=\frac{x+2014}{2013\left(x+1\right)}\)

\(\Leftrightarrow2014x=2013x+2013\)

\(\Leftrightarrow x=2013\)

có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1

=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)

vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015

x=2015

=\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)\)

=\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2106}{2013}-\frac{x-2016}{2012}\)

=\(\left(x-2016\right).\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\)

Mà: \(\frac{1}{2012}>\frac{1}{2015}\)  và \(\frac{1}{2014}< \frac{1}{2013}\)

=>\(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\)  khác \(0\)

Nên: \(x-2016=0\)

=>\(x=2016\)

\(\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}=\frac{x-1}{2014}+\frac{x-1}{2015}\)

\(\Rightarrow\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}-\frac{x-1}{2014}-\frac{x-1}{2015}=0\)

\(\left(x-1\right).\left(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

mà \(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)

=> x - 1 = 0

x = 1

bn có chép sai đề ko z???

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

\(\Leftrightarrow\dfrac{2-x}{2013}+1-x=\dfrac{1-x}{2014}+1-\dfrac{x}{2015}\)

\(\Leftrightarrow\dfrac{2015-x}{2013}-x=\dfrac{2015-x}{2014}-\dfrac{x}{2015}\)

\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}-\dfrac{x}{2015}+x\)

\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}-\dfrac{x}{2015}+1+x-1\)

\(\Leftrightarrow2015-x=0\)

hay x=2015