[tex]\it 6\sqrt5=\sqrt{6^2\cdot5}=\sqrt{36\cdot5}=\sqrt{180}\\ \\ 15=\sqrt{225}\\ \\ \sqrt{180} < \sqrt{225} \Rightarrow 6\sqrt5 < 15 \Rightarrow 6\sqrt5-15 < 0 \Rightarrow |6\sqrt5-15|=-6\sqrt5+15\\ \\ \\ \sqrt{(3\sqrt5-7)^2}=|3\sqrt5-7|\\ \\ 3\sqrt5=\sqrt{3^2\cdot5}=\sqrt{9\cdot5}=\sqrt{45}\\ \\ 7=\sqrt{49}\\ \\ \sqrt{45} < \sqrt{49} \Rightarrow 3\sqrt5 < 7 \Rightarrow 3\sqrt5-7 < 0 \Rightarrow |3\sqrt5-7|=-3\sqrt5+7=7-3\sqrt5[/tex]
[tex]\it Acum,\ \ vom\ \ avea:\\ \\ x=-6\sqrt5+15-2(7-3\sqrt5)=-6\sqrt5+15-14+6\sqrt5=1\\ \\ R\breve aspuns\ \ corect\ \ \bf b)\ Ilona[/tex]
