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....

Bài 1 : Ta có :

\(A=\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)

\(A\sqrt{2}=\sqrt{6x+2\sqrt{6x-1}}+\sqrt{6x-2\sqrt{6x-1}}\)

\(=\sqrt{6x-1+2\sqrt{6x-1}+1}+\sqrt{6x-1-2\sqrt{6x-1}+1}\)

\(=\sqrt{\left(\sqrt{6x-1}+1\right)^2}+\sqrt{\left(\sqrt{6x-1}-1\right)^2}\)

\(=\left|\sqrt{6x-1}+1\right|+\left|\sqrt{6x-1}-1\right|\)

\(=\sqrt{6x-1}+1+\sqrt{6x-1}-1\)

\(=2\sqrt{6x-1}\)

\(\Rightarrow A=\sqrt{2}\left(\sqrt{6x-1}\right)\)

Thay \(x=4+\sqrt{10}\) vào A ta được :

\(A=\sqrt{2}.\sqrt{6\left(4+\sqrt{10}\right)-1}=\sqrt{2}.\sqrt{24+6\sqrt{10}-1}\)

\(=\sqrt{2}.\sqrt{23+6\sqrt{10}}=\sqrt{46+12\sqrt{10}}\)

\(=\sqrt{36+12\sqrt{10}+10}=\sqrt{\left(6+\sqrt{10}\right)^2}=6+\sqrt{10}\)

Vậy \(A=6+\sqrt{10}\) tại \(x=4+\sqrt{10}\)

\(2\le\sqrt{x}+\sqrt{4-x}\le2\sqrt{2}\) (1) (ĐK: \(\left\{{}\begin{matrix}x\ge0\\4-x\ge0\end{matrix}\right.\)\(\Leftrightarrow0\le x\le4\))

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}2\le\sqrt{x}+\sqrt{4-x}\\\sqrt{x}+\sqrt{4-x}\le2\sqrt{2}\end{matrix}\right.\) (\(0\le x\le4\))

\(\Leftrightarrow\left\{{}\begin{matrix}4\le4+2\sqrt{x\left(4-x\right)}\\4+2\sqrt{x\left(4-x\right)}\le8\end{matrix}\right.\) (\(0\le x\le4\))

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x\left(4-x\right)}\ge0\\\sqrt{x\left(4-x\right)}\le2\end{matrix}\right.\)(\(0\le x\le4\))

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(4-x\right)\le4\\0\le x\le4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\0\le x\le4\end{matrix}\right.\) (đpcm)

Ta có:

\(\left\{{}\begin{matrix}\left(2x-5\right)^{2012}\ge0\\\left(3y+4\right)^{2014}\ge0\end{matrix}\right.\forall xy.\)

=> \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\ge0\) \(\forall xy\)

Mà \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}\le0.\)

=> \(\left(2x-5\right)^{2012}+\left(3y+4\right)^{2014}=0\)

=> \(\left(2x-5\right)+\left(3y+4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)

Chúc em học tốt!

Chắc bạn ghi nhầm đề hay sao ấy. 2014 hay 2012 vậy b

Giả sử đề bạn là 2012 thì mình làm nhé.

\(x^4+\sqrt{x^2+2012}=2012\)

\(\Leftrightarrow\left(x^4+x^2+\dfrac{1}{4}\right)=\left(x^2+2012-\sqrt{x^2+2012}+\dfrac{1}{4}\right)\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{2}\right)^2=\left(\sqrt{x^2+2012}-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow x^2+\dfrac{1}{2}=\sqrt{x^2+2012}-\dfrac{1}{2}\)

\(\Leftrightarrow\left(x^2+2012-\sqrt{x^2+2012}+\dfrac{1}{4}\right)=2011,25\)

\(\Leftrightarrow\left(\sqrt{x^2+2012}-\dfrac{1}{2}\right)^2=2011,25\)

Tới đây thì đơn giản rồi. b làm tiếp nhé

Sửa đề: \(\sqrt{2010}-2\sqrt{2012}+\sqrt{2014}< 0\)

Ta có: \(\left(\sqrt{2010}+\sqrt{2014}\right)^2\)

\(=2010+2\sqrt{2010\cdot2014}+2014\)

\(=4024+2\sqrt{\left(2012-2\right)\left(2012+2\right)}\)

\(=2\cdot2012+2\sqrt{2012^2-2^2}\)

\(< 2\cdot2012+2\cdot\sqrt{2012^2}=2\cdot2012+2\cdot2012\)

\(=4\cdot2012=\left(2\sqrt{2012}\right)^2\)

\(\Rightarrow\sqrt{2010}+\sqrt{2014}< 2\sqrt{2012}\)

\(\Leftrightarrow\sqrt{2010}-2\sqrt{2012}+\sqrt{2014}< 0\)

Không đc sửa đề nhé ! Đây là bài chuẩn đấy .

Thế thì sorry nhé, bạn xem lại đề hộ đê

Ta có: \(\sqrt{2012}-2\sqrt{2012}+\sqrt{2014}\)

\(=\sqrt{2014}-\sqrt{2012}>0\)

Cái này hẳn là đề "chuẩn" đấy nhỉ

ko bit

ai giúp mình mình cảm ơn