- Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\ge0\\\sqrt{y-2}=b\ge0\\\sqrt{z-2}=c\ge0\end{matrix}\right.\)

- Đẳng thức đã cho tương đương:

\(a^2+b^2+c^2+3=2a+2b+2c\)

\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-2}=1\\\sqrt{z-2}=1\end{matrix}\right.\)

\(\Rightarrow x=y=z=3\)

\(Q=\sqrt{\left(x-y+1\right)^{2012}}+\sqrt{\left(x-3\right)^{2014}}+\sqrt{\left(y-4\right)^{2016}}\)

\(=\sqrt{1^{2012}}+\sqrt{\left(3-3\right)^{2014}}+\sqrt{\left(3-4\right)^{2016}}\)

\(=\sqrt{1}+\sqrt{0}+\sqrt{1}=2\)

(x-4,5)⁴(x-5,5)⁴ = 0

\(\Rightarrow\left[{}\begin{matrix}x-4,5=0\\x-5,5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4,5\\x=5,5\end{matrix}\right.\)

Điều kiện: \(x\ge1\)

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

$\sqrt{x+2\sqrt{x-1}}=2$$\iff \sqrt{\left(x-1\right)+2.\sqrt{x-1}.1+1}=2$

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

$\iff \sqrt{x-1}+1=2$

$\iff$$\sqrt{x-1}=1$

$\iff {\left(\sqrt{x-1}\right)}^2=1^2$

$\iff x-1=1$

$\iff x=2$

\(\sqrt{x^2}=7\Rightarrow\left|x\right|=7\Rightarrow x=\pm7\)

x=7

Với \(x\le-3\Rightarrow x+3\le0\)

Ta có \(3x+\sqrt{9+6x+x^2}=3x+\sqrt{\left(x+3\right)^2}=3x+\left|x+3\right|=3x-x-3=2x-3\)