Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 405

а) $$ 3 \sqrt{\frac{1}{3}}=\sqrt{3^2 \square \frac{1}{3}}=\sqrt{\frac{9 \square}{3}}=\sqrt{\frac{9}{3}} $$ = $$\sqrt{3}$$
б) $$ 2 \sqrt{\frac{3}{4}}=\sqrt{2^2-\frac{3}{4}}=\sqrt{\frac{4 B}{4}}=\sqrt{\frac{4 B \div 4}{4 \div 4}} $$ = $$\sqrt{3}$$
в) 1/3$$\sqrt{18}$$ = $$ \sqrt{\left(\frac{1}{3}\right)^2 \square 8}=\sqrt{\frac{1}{9} \square 8}=\sqrt{\frac{118}{9}}=\sqrt{\frac{118 \div 9}{9 \div 9}} $$ = $$\sqrt{2}$$
г) −10$$\sqrt{0,02}$$ = −$$\sqrt{10² ⋅ 0,02}$$ = −$$\sqrt{100 ⋅ 0,02}$$ = −$$\sqrt{2}$$
д) $$ 5 \sqrt{\frac{a}{5}}=\sqrt{5^2 \frac{a}{5}}=\sqrt{25 \frac{a}{5}}=\sqrt{\frac{25 \square a}{5}}=\sqrt{\frac{25 \square a \div 5}{5 \div 5}} $$ = $$\sqrt{5a}$$
е) −1/2$$\sqrt{12x}$$ = $$ -\sqrt{\left(\frac{1}{2}\right)^2 \square 2 x}=-\sqrt{\frac{1 \square 2 x}{4}}=-\sqrt{\frac{12 x \div 4}{4 \div 4}} $$ = −$$\sqrt{3x}$$
ж) −0,1$$\sqrt{1,2a}$$ = −$$\sqrt{0,1² ⋅ 1,2a}$$ = −$$\sqrt{0,01 ⋅ 1,2a}$$ = −$$\sqrt{0,012a}$$
з) −1/3$$\sqrt{0,9a}$$ = $$ -\sqrt{\left(\frac{1}{3}\right)^2 \frac{9 a}{10}}=-\sqrt{\frac{19 a}{9 \square 0}}=-\sqrt{\frac{9 a \div 9}{9 \square 0 \div 9}}=-\sqrt{\frac{a}{10}} $$
и) −6$$\sqrt{6b}$$ = −$$\sqrt{6² ⋅ 6b}$$ = −$$\sqrt{36 ⋅ 6b}$$ = −$$\sqrt{216b}$$