Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 505

а) $$ 15 \sqrt{\frac{2}{5}}-\sqrt{160}=\sqrt{225 \frac{2}{5}} $$ − $$\sqrt{16 ⋅ 10}$$ = $$\sqrt{45}$$ − $$\sqrt{16 ⋅ 10}$$ = $$\sqrt{45 ⋅ 2}$$ − $$\sqrt{16 ⋅ 10}$$ = $$\sqrt{90}$$ − $$\sqrt{16 ⋅ 10}$$ = $$\sqrt{9 ⋅ 10}$$ − $$\sqrt{16 ⋅ 10}$$ = 3$$\sqrt{10}$$ − 4$$\sqrt{10}$$ = −$$\sqrt{10}$$
б) $$\sqrt{135}$$ + 10$$\sqrt{0,6}$$ = $$\sqrt{9 ⋅ 15}$$ + $$\sqrt{100 ⋅ 0,6}$$ = $$\sqrt{9 ⋅ 15}$$ + $$\sqrt{60}$$ = $$\sqrt{9 ⋅ 15}$$ + $$\sqrt{4 ⋅ 15}$$ = 3$$\sqrt{15}$$ + 2$$\sqrt{15}$$ = 5$$\sqrt{15}$$
в) $$ 6 \sqrt{1 \frac{1}{3}}-\sqrt{27}=6 \sqrt{\frac{4}{3}}-\sqrt{27}=\sqrt{36 \frac{4}{3}} $$ − $$\sqrt{27}$$ = $$\sqrt{12 ⋅ 4}$$ − $$\sqrt{27}$$ = $$\sqrt{48}$$ − $$\sqrt{27}$$ = $$\sqrt{16 ⋅ 3}$$ − $$\sqrt{9 ⋅ 3}$$ = 4$$\sqrt{3}$$ − 3$$\sqrt{3}$$ = $$\sqrt{3}$$
г) 0,5$$\sqrt{24}$$ + 10$$ \sqrt{\frac{3}{8}} $$ = $$\sqrt{0,25 ⋅ 24}$$ + 10$$ \sqrt{\frac{6}{16}} $$ = $$\sqrt{3}$$ + 10$$ \sqrt{\frac{1}{16} \square 6} $$ = $$\sqrt{6}$$ + 10 ⋅ 1/4$$\sqrt{6}$$ = $$\sqrt{6}$$ + 2,5$$\sqrt{6}$$ = 3,5$$\sqrt{6}$$