$1=\sqrt{1}$
$\sqrt{1}<\sqrt{2}$
$1<\sqrt{2}$
тогда:
$\sqrt{(1-\sqrt{2}{)}^2}=|1-\sqrt{2}|=-1+\sqrt{2}=\sqrt{2}-1$

$\sqrt{6}<\sqrt{7}$
тогда:
$\sqrt{(\sqrt{6}-\sqrt{7}{)}^2}=|\sqrt{6}-\sqrt{7}|=-\sqrt{6}+\sqrt{7}=\sqrt{7}-\sqrt{6}$

$2\sqrt{5}=\sqrt{4\ast 5}=\sqrt{20}$
$3=\sqrt{9}$
$\sqrt{20}>\sqrt{9}$
$2\sqrt{5}>3$
тогда:
$\sqrt{(2\sqrt{5}-3{)}^2}=|2\sqrt{5}-3|=2\sqrt{5}-3$

$2=\sqrt{4}$
$\sqrt{3}<\sqrt{4}$
$\sqrt{3}<2$
$3=\sqrt{9}$
$\sqrt{9}>\sqrt{3}$
$3>\sqrt{3}$
тогда:
$\sqrt{(\sqrt{3}-2{)}^2}+\sqrt{(3-\sqrt{3}{)}^2}=|\sqrt{3}-2|+|3-\sqrt{3}|=2-\sqrt{3}+3-\sqrt{3}=5-2\sqrt{3}$