Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 429

а) $$ \frac{x}{x+\sqrt{y}}=\frac{x(x-\sqrt{y})}{(x+\sqrt{y})(x-\sqrt{y})}=\frac{x(x-\sqrt{y})}{x^2-y} $$
б) $$ \frac{b}{a-\sqrt{b}}=\frac{b(a+\sqrt{b})}{(a-\sqrt{b})(a+\sqrt{b})}=\frac{b(a+\sqrt{b})}{a^2-b} $$
в) $$ \frac{4}{\sqrt{10}-\sqrt{2}}=\frac{4(\sqrt{10}+\sqrt{2})}{(\sqrt{10}-\sqrt{2})(\sqrt{10}+\sqrt{2})}=\frac{4(\sqrt{10}+\sqrt{2})}{\sqrt{10}^2-\sqrt{2}^2}=\frac{14(\sqrt{10}+\sqrt{2})}{10-2}= $$ $$ \frac{4(\sqrt{10}+\sqrt{2})}{8}=\frac{4(\sqrt{10}+\sqrt{2}) \div 4}{8 \div 4}=\frac{\sqrt{10}+\sqrt{2}}{2} $$
г) $$ \frac{12}{\sqrt{3}+\sqrt{6}}=\frac{12(\sqrt{3}-\sqrt{6})}{(\sqrt{3}+\sqrt{6})(\sqrt{3}-\sqrt{6})}=\frac{12(\sqrt{3}-\sqrt{6})}{\sqrt{3}^2-\sqrt{6}^2}=\frac{12(\sqrt{3}-\sqrt{6})}{3-6}= $$ $$ \frac{12(\sqrt{3}-\sqrt{6})}{-3}=\frac{12(\sqrt{3}-\sqrt{6}) \div(-3)}{-3 \div(-3)} $$ = −4($$\sqrt{3}$$ − $$\sqrt{6}$$) = 4$$\sqrt{6}$$ − 4$$\sqrt{3}$$
д) $$ \frac{9}{3-2 \sqrt{2}}=\frac{9(3+2 \sqrt{2})}{(3-2 \sqrt{2})(3+2 \sqrt{2})}=\frac{9(3+2 \sqrt{2})}{3^2-(2 \sqrt{2})^2}=\frac{9(3+2 \sqrt{2})}{9-8}= $$ $$ \frac{9(3+2 \sqrt{2})}{1} $$ = 9 ⋅ (3 + 2$$\sqrt{2}$$) = 27 + 18$$\sqrt{2}$$
е) $$ \frac{14}{1+5 \sqrt{2}}=\frac{14(1-5 \sqrt{2})}{(1+5 \sqrt{2})(1-5 \sqrt{2})}=\frac{14(1-5 \sqrt{2})}{1^2-(5 \sqrt{2})^2}=\frac{14(1-5 \sqrt{2})}{1-50}= $$ $$ \frac{14(1-5 \sqrt{2})}{49}=\frac{14(1-5 \sqrt{2}) \div 7}{49 \div 7}=\frac{2(1-5 \sqrt{2})}{7} $$