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

x^2=t≥0

t^2+√(t+2014)=2014

√(t+2014)=a; a≥√2014

a^2=t+2014(1)

t^2+a=2014(2)

(1)-(2)

(a-t)(a+t)=-(a-t)

th1

a=t; =>t≥√2014

(2)=>t^2+t-2014=0

∆=1+4.2014

t=(√(1+4.2014)-1)/2

x=±√t

th2

a+t=-1

a=-t-1=>0≤t≤√(2014)-1

t^2-t-2015=0

(tu gq tiep)

?

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