Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 428

а) $$ \frac{1}{\sqrt{5}-2}=\frac{1(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}=\frac{\sqrt{5}+2}{\sqrt{5}^2-2^2}=\frac{\sqrt{5}+2}{5-4}=\frac{\sqrt{5}+2}{1} $$ = $$\sqrt{5}$$ + 2
2 < $$\sqrt{5}$$ < 3 2 + 2 < $$\sqrt{5}$$ < 3 + 2 4 < $$\sqrt{5}$$ < 5
б) $$ \frac{2}{\sqrt{5}-\sqrt{3}}=\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2(\sqrt{5}+\sqrt{3})}{\sqrt{5}^2-\sqrt{3}^2}=\frac{2(\sqrt{5}+\sqrt{3})}{5-3} $$ = = $$ \frac{2(\sqrt{5}+\sqrt{3})}{2}=\frac{2(\sqrt{5}+\sqrt{3}) \div 2}{2 \div 2} $$ = $$\sqrt{5}$$ + $$\sqrt{3}$$
2 < $$\sqrt{5}$$ < 3 1 < $$\sqrt{3}$$ < 2 2 + 1 < $$\sqrt{5}$$ + $$\sqrt{3}$$ < 3 + 2 3 < $$\sqrt{5}$$ + $$\sqrt{3}$$ < 5 $$\sqrt{5}$$ ≈ 2,2; $$\sqrt{3}$$ ≈ 1,7 $$\sqrt{5}$$ + $$\sqrt{3}$$ ≈ 2,2 + 1,7 = 3,9 3 < $$\sqrt{5}$$ + $$\sqrt{3}$$ < 4
в) $$ \frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3(\sqrt{10}-\sqrt{7})}{(\sqrt{10}+\sqrt{7})(\sqrt{10}-\sqrt{7})}=\frac{3(\sqrt{10}-\sqrt{7})}{\sqrt{10}^2-\sqrt{7}^2}=\frac{3(\sqrt{10}-\sqrt{7})}{10-7} $$ = = $$ \frac{3(\sqrt{10}-\sqrt{7})}{3}=\frac{3(\sqrt{10}-\sqrt{7}) \div 3}{3 \div 3}= $$ = $$\sqrt{10}$$ − $$\sqrt{7}$$
3 < $$\sqrt{10}$$ < 4 2 < $$\sqrt{7}$$ < 3 −3 < −$$\sqrt{7}$$ < −2 3 − 3 < $$\sqrt{10}$$ − $$\sqrt{7}$$ < 4 − 2 0 < $$\sqrt{10}$$ − $$\sqrt{7}$$ < 2 $$\sqrt{10}$$ ≈ 3,2; $$\sqrt{7}$$ ≈ 2,6 $$\sqrt{10}$$ − $$\sqrt{7}$$ ≈ 3,2 − 2,6 = 0,6 0 < $$\sqrt{10}$$ − $$\sqrt{7}$$ < 1
г) $$ \frac{5+3 \sqrt{3}}{\sqrt{3}+2}=\frac{(5+3 \sqrt{3})(\sqrt{3}-2)}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{5 \sqrt{3}-10+9-6 \sqrt{3}}{\sqrt{3}^2-2^2}=\frac{-1-\sqrt{3}}{3-4}= $$ = = $$ \frac{-1(1+\sqrt{3})}{-1}=\frac{-1(1+\sqrt{3}) \div(-1)}{-1 \div(-1)} $$ = 1 + $$\sqrt{3}$$
1 < $$\sqrt{3}$$ < 2 1 + 1 < 1 + $$\sqrt{3}$$ < 2 + 1 2 < 1 + $$\sqrt{3}$$ < 3