đặt a = x^2 
b = -căn(x^2 + 2014) 
=> a^2 - b = 2014 
và :b^2 = a+2014 
=> (a-b).(a+b+1) = 0 

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....

\(\Leftrightarrow x^4\left(\sqrt{x+3}-2\right)+2014\left(x-1\right)=0\)

\(\Leftrightarrow x^4\cdot\frac{x-1}{\sqrt{x+3}+2}+2014\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x^4}{\sqrt{x+3}+2}+2014\right)=0\)

Dễ thấy \(\left(\frac{x^4}{\sqrt{x+3}+2}+2014\right)\ne0\)

\(\Rightarrow x=1\)

Học tốt

?

khó

mik ko bít phantuananh a

\(\sqrt{x+123234048-22012\sqrt{x+2102012}}\)

\(=\sqrt{x+2102012-2.11006\sqrt{x+2102012}+121132036}\)

\(=\sqrt{\left(\sqrt{x+2102012}-11006\right)^2}\)

\(=\left|\sqrt{x+2102012}-11006\right|\)

\(\sqrt{x+103426368-20132\sqrt{x+2102012}}\)

\(=\sqrt{x+2102012-2.10066.\sqrt{x+2102012}+101324356}\)

\(=\sqrt{\left(\sqrt{x+2102012}-10066\right)^2}\)

\(=\left|\sqrt{x+2102012}-10066\right|\) 

Bạn thế vào pt rồi chia trường hợp