[tex]\it a)\ 5\sqrt7=\sqrt{5^2\cdot7}=\sqrt{25\cdot7}=\sqrt{175}\\ \\ 6\sqrt5=\sqrt{6^2\cdot5}=\sqrt{36\cdot5}=\sqrt{180}\\ \\ \sqrt{175}<\sqrt{180} \Rightarrow 5\sqrt7<6\sqrt5 \Rightarrow 5\sqrt7-6\sqrt5<0 \Rightarrow\\ \\ \Rightarrow |5\sqrt7-6\sqrt5| =-(5\sqrt7-6\sqrt5) =6\sqrt5-5\sqrt7\\ \\ \\ b)\ \dfrac{\sqrt{63}}{3}=\dfrac{\sqrt{9\cdot7}}{3}=\dfrac{3\sqrt7}{3}=\sqrt7[/tex]
[tex]\it c)\ \ 3\sqrt7=\sqrt{3^2\cdot7}=\sqrt{9\cdot7}=\sqrt{63}>\sqrt{49} \Rightarrow \sqrt{49}<\sqrt{63} \Rightarrow 7<3\sqrt7\\ \\ 7\sqrt2=\sqrt{7^2\cdot2}=\sqrt{49\cdot2}=\sqrt{98}<\sqrt{100} \Rightarrow 7\sqrt2<10\\ \\ \\ Deci, vom\ avea:\\ \\ 7<3\sqrt7<x<7\sqrt2<10 \Rightarrow 7<x<10 \Rightarrow x\in\{8,\ 9\}[/tex]
