Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 432

15$$\sqrt{3}$$ − 4$$\sqrt{2}$$
6 − $$\sqrt{12}$$ = 6 − $$\sqrt{4 ⋅ 3}$$ = 6 − 2$$\sqrt{3}$$
$$\sqrt{80}$$ − 5$$\sqrt{3}$$ = $$\sqrt{16 ⋅ 5}$$ − 5$$\sqrt{3}$$ = 4$$\sqrt{5}$$ − 5$$\sqrt{3}$$
$$\sqrt{75}$$ − 4$$\sqrt{5}$$ = $$\sqrt{25 ⋅ 3}$$ − 4$$\sqrt{5}$$ = 5$$\sqrt{3}$$ − 4$$\sqrt{5}$$
$$ \frac{1}{2 \sqrt{3}-6}=\frac{1(2 \sqrt{3}+6)}{(2 \sqrt{3}-6)(2 \sqrt{3}+6)}=\frac{2 \sqrt{3}+6}{(2 \sqrt{3})^2-6^2}=\frac{2 \sqrt{3}+6}{2^2 \square \sqrt{3}^2-6^2}= $$ = $$ \frac{2 \sqrt{3}+6}{4 \sqrt{3}-36}=\frac{2 \sqrt{3}+6}{12-36}=\frac{2(\sqrt{3}+3) \div 2}{-24 \div 2}=-\frac{\sqrt{3}+3}{12} $$ = $$ \frac{1}{\sqrt{675}-\sqrt{32}}=\frac{1(\sqrt{675}+\sqrt{32})}{(\sqrt{675}-\sqrt{32})(\sqrt{675}+\sqrt{32})}=\frac{\sqrt{225}+\sqrt{162}}{\sqrt{675}^2-\sqrt{32}^2}= $$ = $$ \frac{15 \sqrt{3}+4 \sqrt{2}}{675-32}=\frac{15 \sqrt{3}+4 \sqrt{2}}{643} $$
4$$\sqrt{5}$$ − 5$$\sqrt{3}$$ + 5$$\sqrt{3}$$ − 4$$\sqrt{5}$$ = 0, значит числа $$\sqrt{80}$$ − 5$$\sqrt{3}$$, $$\sqrt{75}$$ − 4$$\sqrt{5}$$ противоположные
15$$\sqrt{3}$$ − 4$$\sqrt{2}$$ ⋅ $$ \frac{15 \sqrt{3}+4 \sqrt{2}}{643}=\frac{(15 \sqrt{3}-4 \sqrt{2})(15 \sqrt{3}+4 \sqrt{2})}{643}=\frac{(15 \sqrt{3})^2-(4 \sqrt{2})^2}{643}= $$ = $$ \frac{15^2(\sqrt{3})^2-4^2(\sqrt{2})^2}{643} $$ = 225 ⋅ 3 − 16 ⋅ 2/643 = 675 − 32/643 = 643/643 = 1, значит числа 15$$\sqrt{3}$$ − 4$$\sqrt{2}$$, $$ \frac{1}{\sqrt{675}-\sqrt{32}} $$ взаимно обратные