Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 426

а) $$ \frac{4}{\sqrt{3}+1}=\frac{4(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4(\sqrt{3}-1)}{\sqrt{3}^2-1^2}=\frac{4(\sqrt{3}-1) \div 2}{2 \div 2}=\frac{2(\sqrt{3}-1)}{1} $$ = 2($$\sqrt{3}$$ − 1) = 2$$\sqrt{3}$$ − 2
б) $$ \frac{1}{1-\sqrt{2}}=\frac{1(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}=\frac{1+\sqrt{2}}{1^2-\sqrt{2}^2}=\frac{1+\sqrt{2}}{1-2}=\frac{1+\sqrt{2}}{-1} $$ = −1 − $$\sqrt{3}$$
в) $$ \frac{1}{\sqrt{x}-\sqrt{y}}=\frac{1(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}^2-\sqrt{y}^2}=\frac{\sqrt{x}+\sqrt{y}}{x-y} $$
г) $$ \frac{a}{\sqrt{a}+\sqrt{b}}=\frac{a(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}=\frac{a(\sqrt{a}-\sqrt{b})}{\sqrt{a}^2-\sqrt{b}^2}=\frac{a(\sqrt{a}-\sqrt{b})}{a-b} $$
д) $$ \frac{33}{7-3 \sqrt{3}}=\frac{33(7+3 \sqrt{3})}{(7-3 \sqrt{3})(7+3 \sqrt{3})}=\frac{33(7+3 \sqrt{3})}{7^2-(3 \sqrt{3})^2}=\frac{33(7+3 \sqrt{3})}{49-9 \sqrt{3}}=\frac{33(7+3 \sqrt{3})}{49-27}= $$ $$ \frac{33(7+3 \sqrt{3})}{22}=\frac{33(7+3 \sqrt{3}) \div 11}{22 \div 11}=\frac{3(7+3 \sqrt{3})}{2} $$
е) $$ \frac{15}{2 \sqrt{5}+5}=\frac{15(2 \sqrt{5}-5)}{(2 \sqrt{5}+5)(2 \sqrt{3}-5)}=\frac{15(2 \sqrt{5}-5)}{(2 \sqrt{5})^2-5^2}=\frac{15(2 \sqrt{5}-5)}{2^2-\sqrt{5}^2-5^2}=\frac{15(2 \sqrt{5}-5)}{45-25}= $$ $$ \frac{15(2 \sqrt{5}-5)}{-5}=\frac{15(2 \sqrt{5}-5) \div 5}{-5 \div 5}=\frac{3(2 \sqrt{5}-5)}{-1} $$ = −3(2$$\sqrt{5}$$ − 5) = 6$$\sqrt{5}$$ + 15 = 15 − 6$$\sqrt{5}$$