679* Вправа 679Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 197

679. Звільніться від ірраціональності в знаменнику дробу:

1) ${\displaystyle \frac{1}{1+\sqrt{3}-\sqrt{2}}=}$${\displaystyle \frac{1+\sqrt{3}+\sqrt{2}}{{(1+\sqrt{3})}^2-2}=}$${\displaystyle \frac{1+\sqrt{3}+\sqrt{2}}{2(1+\sqrt{3})}=}$${\displaystyle \frac{(1+\sqrt{3}+\sqrt{2})(\sqrt{3}-1)}{2\cdot 2}=}$${\displaystyle \frac{2+\sqrt{6}-\sqrt{2}}{4}}$;

2) ${\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=}$${\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-5}=}$${\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}=}$${\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}}$;

3) ${\displaystyle \frac{1}{\sqrt{3}+\sqrt{7}-\sqrt{10}}=}$${\displaystyle \frac{\sqrt{3}+\sqrt{7}+\sqrt{10}}{{(\sqrt{3}+\sqrt{7})}^2-10}=}$${\displaystyle \frac{\sqrt{3}+\sqrt{7}+\sqrt{10}}{2\sqrt{21}}=}$${\displaystyle \frac{3\sqrt{7}+7\sqrt{3}+\sqrt{210}}{42}}$;

4) ${\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{6}}=}$${\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{{(\sqrt{2}+\sqrt{3})}^2-6}=}$${\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}=}$${\displaystyle \frac{(\sqrt{2}+\sqrt{3}-\sqrt{6})(2\sqrt{6}+1)}{24-1}=}$${\displaystyle \frac{5\sqrt{3}+7\sqrt{2}-\sqrt{6}-12}{23}}$.