Ta có :
2|3x-6|+(x-2)^2 = 0
=> 0+0=0
=>2|3x-6 |=0
=>3x-6 =0
=> 0+6= 3x
=>x=2
Ta có
(x+2)^2=0
x^2+4 =0
=> x^2 =4 => x =2

$2\left| 3x-6 \right|+{{\left( x-2 \right)}^{2}}=\text{ }0$
Ta có:
$\left\{ \begin{align} & 2\left| 2x-6 \right|\ge 0 \\ & {{\left( x-2 \right)}^{2}}\ge\\\end{align} \right.$$\Rightarrow 2\left| 3x-6 \right|+{{\left( x-2 \right)}^{2}}=\text{ }0$
$\Leftrightarrow \left\{ \begin{align} & \left| 3x-6 \right|=0 \\ & {{\left( x-2 \right)}^{2}}=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & x=2 \\ & x=2 \\ \end{align} \right.$$\Leftrightarrow x=2$

Ta có :
2|3x-6|+(x-2)^2 = 0
=> 0+0=0
=>2|3x-6 |=0
=>3x-6 =0
=> 0+6= 3x
=>x=2
Ta có
(x+2)^2=0
x^2+4 =0
=> x^2 =4 => x =2