[tex]\it \bf{6}.[/tex]
[tex]\it a)\ 7\sqrt3-\sqrt{12}=7\sqrt3-\sqrt{4\cdot3}=7\sqrt3-2\sqrt3=5\sqrt3\\ \\ c)\sqrt{48}-\sqrt{12}=\sqrt{16\cdot3}-\sqrt{4\cdot3}=4\sqrt3-2\sqrt3=2\sqrt3[/tex]
[tex]\bf7.[/tex]
[tex]\it a)\ \sqrt8+\sqrt{}32-\sqrt{50}=\sqrt{4\cdot2}+\sqrt{16\cdot2}-\sqrt{25\cdot2}=2\sqrt2+4\sqrt2-5\sqrt2=\sqrt2\\ \\ c)\ \sqrt{28}+\sqrt{63}-5\sqrt7=2\sqrt7+3\sqrt7-5\sqrt7=0[/tex]
[tex]\bf8.[/tex]
[tex]\it b)\ \sqrt{162}+\sqrt8-\sqrt{288}+\sqrt{18}=\sqrt{81\cdot2}+\sqrt{4\cdot2}-\sqrt{144\cdot2}+\sqrt{9\cdot2}=\\ \\ =9\sqrt2+2\sqrt2-12\sqrt2+3\sqrt2=14\sqrt2-12\sqrt2=2\sqrt2[/tex]
[tex]\bf9.[/tex]
[tex]\it b)\ \dfrac{^{2)}\sqrt{20}}{3}+\dfrac{^{3)}\sqrt{45}}{2}+\dfrac{13\sqrt5}{6}=\dfrac{2\sqrt{4\cdot5}+3\sqrt{9\cdot5}-13\sqrt5}{6}=\\ \\ \\ =\dfrac{4\sqrt5+9\sqrt5-13\sqrt5}{6}=\dfrac{0}{6}=0[/tex]
[tex]\bf10.[/tex]
[tex]\it a)\ \sqrt{0,8}+2\sqrt{1,8}-8\sqrt{0,2}=\sqrt{0,2}(\sqrt4+2\sqrt9-8\sqrt1)=\\ \\ \\ = \sqrt{0,2}(2+6-8)=\sqrt{0,2}\cdot0=0[/tex]
