[tex]\it 2)\ 9>8 \Rightarrow \sqrt9>\sqrt8 \Rightarrow 3>\sqrt{4\cdot2} \Rightarrow 3>2\sqrt2 \Rightarrow 3-2\sqrt2>0 \Rightarrow \\ \\ \Rightarrow |3-2\sqrt2|=3-2\sqrt2\\ \\ \sqrt{(2\sqrt2-3)^2}=\sqrt{(3-2\sqrt2)^2}=|3-2\sqrt2|=3-2\sqrt2[/tex]
[tex]\it 3)\ \ a\cdotb=\sqrt{2-2\sqrt2}\cdot\sqrt{2+\sqrt2}=\sqrt{(2-\sqrt2)(2+\sqrt2)}=\\ \\ =\sqrt{2^2-(\sqrt2)^2}=\sqrt{4-2}=\sqrt2=\sqrt2[/tex]
