Упр.105 Вариант 2 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)
- Найдите значение выражения:
1) $$\frac{6}{7-3\sqrt{5}}-\frac{6}{7+3\sqrt{5}}$$;
2) $$\frac{1}{\sqrt{5+\sqrt{12}}-1}-\frac{1}{\sqrt{5+\sqrt{12}}+1}$$;
3) $$\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}-\sqrt{13}}+\frac{\sqrt{17}-\sqrt{13}}{\sqrt{17}+\sqrt{13}}$$.
1) $$\frac{6}{7-3\sqrt5}-\frac{6}{7+3\sqrt5}=\frac{6(7+3\sqrt5)-6(7-3\sqrt5)}{(7-3\sqrt5)(7+3\sqrt5)}$$
$$=\frac{42+18\sqrt5-42+18\sqrt5}{49-9\cdot 5}=\frac{36\sqrt5}{4}=9\sqrt5.$$
2) $$\frac{1}{\sqrt{5+\sqrt{12}}-1}-\frac{1}{\sqrt{5+\sqrt{12}}+1}$$
$$=\frac{\left(\sqrt{5+\sqrt{12}}+1\right)-\left(\sqrt{5+\sqrt{12}}-1\right)}{\left(\sqrt{5+\sqrt{12}}-1\right)\left(\sqrt{5+\sqrt{12}}+1\right)}$$
$$=\frac{2}{(5+\sqrt{12})-1}=\frac{2}{4+\sqrt{12}}=\frac{2}{4+2\sqrt3}=\frac{1}{2+\sqrt3}$$
$$=\frac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}=2-\sqrt3.$$
3) $$\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}-\sqrt{13}}+\frac{\sqrt{17}-\sqrt{13}}{\sqrt{17}+\sqrt{13}}$$
$$=\frac{(\sqrt{17}+\sqrt{13})^2+(\sqrt{17}-\sqrt{13})^2}{(\sqrt{17}-\sqrt{13})(\sqrt{17}+\sqrt{13})}$$
$$=\frac{17+2\sqrt{17\cdot 13}+13+17-2\sqrt{17\cdot 13}+13}{17-13}=\frac{60}{4}=15.$$
Ответ: 1) $$9\sqrt5$$; 2) $$2-\sqrt3$$; 3) $$15$$.
