$a\text{)}\sqrt{4x^2-12x+9}-2\sqrt{x^2}+3\left|x+2\right|=$  

$=\sqrt{{\left(2x-3\right)}^2}-2\sqrt{x^2}+3\left|x+2\right|=\left|2x-3\right|-2\left|x\right|+3\left|x+2\right|$  

Wiemy, że:

$x\in \left(-\infty ,-2\right)$  

zatem:  

$\left|2x-3\right|-2\left|x\right|+3\left|x+2\right|=-\left(2x-3\right)+2x+3\cdot \left(-\left(x+2\right)\right)=$  

$=-2x+3+2x-3x-6=-3x-3$  

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$b\text{)}6\left|-x\right|-\left|-4x-4\right|-\sqrt{25-10x+x^2}=$  

$=6\left|-x\right|-\left|-4x-4\right|-\sqrt{{\left(5-x\right)}^2}=6\left|-x\right|-\left|-4x-4\right|-\left|5-x\right|$  

Wiemy, że:

$x\in \left(0,5\right)$  

zatem:

$6\left|-x\right|-\left|-4x-4\right|-\left|5-x\right|=6\cdot x-\left(4x+4\right)-\left(5-x\right)=$  

$=6x-4x-4-5+x=3x-9$  

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$c\text{)}\frac{\sqrt{x^2-12x+36}-\sqrt{4x^2-12x+9}}{x^2-9}=$  

$=\frac{\sqrt{{\left(x-6\right)}^2}-\sqrt{{\left(2x-3\right)}^2}}{x^2-9}=\frac{\left|x-6\right|-\left|2x-3\right|}{x^2-9}$  

Wiemy, że:

$x\in \left(6,+\infty \right)$  

zatem:

$\frac{\left|x-6\right|-\left|2x-3\right|}{x^2-9}=\frac{x-6-\left(2x-3\right)}{x^2-9}=\frac{x-6-2x+3}{x^2-9}=$  

$=\frac{-x-3}{\left(x-3\right)\left(x+3\right)}=-\frac{x+3}{\left(x-3\right)\left(x+3\right)}=-\frac{1}{x-3}$  

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$d\text{)}\frac{\left|x-3\right|\cdot \sqrt{x^2-2x+1}-\sqrt{4x^2-24x+36}}{{\left(x-1\right)}^2-4}=$  

$=\frac{\left|x-3\right|\cdot \sqrt{{\left(x-1\right)}^2}-\sqrt{{\left(2x-6\right)}^2}}{{\left(x-1\right)}^2-4}=\frac{\left|x-3\right|\cdot \left|x-1\right|-\left|2x-6\right|}{{\left(x-1\right)}^2-4}$  

Wiemy, że:

$x\in \mathbb{R}-\left\{-1,3\right\}$  

zatem:

$\frac{\left|x-3\right|\cdot \left|x-1\right|-\left|2x-6\right|}{{\left(x-1\right)}^2-4}=\frac{|x-3|\cdot |x-1|-2|x-3|}{{\left|x-1\right|}^2-2^2}=$  

$=\frac{|x-3|\cancel{\left(|x-1|-2\right)}}{\cancel{\left(|x-1|-2\right)}\left(|x-1|+2\right)}=\frac{\left|x-3\right|}{\left|x-1\right|+2}$