Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)

\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)

\(\Leftrightarrow x=2014\)

Vậy \(x=2014\)

\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)

\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)

Vậy x = 2014

\(x=\dfrac{\dfrac{2013}{1}+\dfrac{2012}{2}+......+\dfrac{2}{2012}+\dfrac{1}{2013}}{\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{2014}}\)

\(=\dfrac{\left(\dfrac{2012}{2}+1\right)+\left(\dfrac{2011}{3}+1\right)+......+\left(\dfrac{1}{2013}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{2014}}\)

\(=\dfrac{\dfrac{2014}{2}+\dfrac{2014}{3}+......+\dfrac{2014}{2013}+\dfrac{2014}{2014}}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{2014}}\)

\(=\dfrac{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{2014}}\)

=> x = 2014

a: \(\Leftrightarrow x+2016=0\)

hay x=-2016

b: \(\Leftrightarrow x-100=0\)

hay x=100

\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)

\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)

\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)

\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)

\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

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 :

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

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

\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)

\(\dfrac{x-3}{2012}+\dfrac{x-2}{2013}=\dfrac{x-2013}{2}+\dfrac{x-2012}{3}\)(mk nghĩ đề như thế này)

\(\Leftrightarrow\dfrac{x-3}{2012}-1+\dfrac{x-2}{2013}-1=\dfrac{x-2013}{2}-1+\dfrac{x-2012}{3}-1\)

\(\Leftrightarrow\dfrac{x-2015}{2012}+\dfrac{x-2015}{2013}=\dfrac{x-2015}{2}+\dfrac{x-2015}{3}\)

\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x=2015\)(vì \(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\))

\(\dfrac{x-3}{2012}+\dfrac{x-2}{2013}=\dfrac{x-2013}{2}+\dfrac{x-2015}{3}\\ \Leftrightarrow\left(\dfrac{x-3}{2012}-1\right)+\left(\dfrac{x-2}{2013}-1\right)=\left(\dfrac{x-2013}{2}-1\right)+\left(\dfrac{x-2015}{3}-1\right)\\ \Leftrightarrow\dfrac{x-2018}{2012}+\dfrac{x-2018}{2013}-\dfrac{x-2018}{2}-\dfrac{x-2018}{3}=0\\ \Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\\ \Leftrightarrow x-2018=0\left(\text{Vì }\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\right)\\ x=2018\)

Vậy phương trình có nghiệm \(x=2018\)

\(\dfrac{x+3}{2011}+\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\ge\dfrac{3x}{2014}\)

\(\dfrac{x+3}{2011}+1+\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\ge\dfrac{3x}{2014}+3\)

\(\dfrac{x+2014}{2011}+\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\ge3\left(\dfrac{x+2014}{2014}\right)\)

\(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{3}{2014}\right)\ge0\)

Mà \(\left(\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{3}{2014}\right)>0\) (bạn có thể chứng minh nếu thích )

Nên \(x+2014\ge0\)

\(\Leftrightarrow x\ge-2014\)

Vậy

\(\Leftrightarrow\dfrac{5-x^2}{2012}=\dfrac{4-x^2}{2013}+1-\dfrac{x^2-3}{2014}\)

\(\Leftrightarrow\dfrac{5-x^2}{2012}+1=\dfrac{4-x^2}{2013}+1+\dfrac{3-x^2}{2014}+1\)

\(\Leftrightarrow2017-x^2=0\)

hay \(x\in\left\{\sqrt{2017};-\sqrt{2017}\right\}\)