664 Вправа 664Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 195

664. Звільніться від ірраціональності в знаменнику дробу:

1) ${\displaystyle \frac{1^{\mathrm{\setminus }\left(1-\sqrt{5}\right)}}{1+\sqrt{5}}=}$${\displaystyle \frac{1-\sqrt{5}}{1^2-{\left(\sqrt{5}\right)}^2}=}$${\displaystyle \frac{1-\sqrt{5}}{-4}=}$${\displaystyle \frac{\sqrt{5}-1}{4}}$;

2) ${\displaystyle \frac{3^{\mathrm{\setminus }\left(\sqrt{7}+2\right)}}{\sqrt{7}-2}=}$${\displaystyle \frac{3(\sqrt{7}+2)}{{\left(\sqrt{7}\right)}^2-2^2}=}$${\displaystyle \frac{3(\sqrt{7}+2)}{3}=}$${\displaystyle \sqrt{7}+2}$;

3) ${\displaystyle \frac{2^{\mathrm{\setminus }\left(\sqrt{5}+\sqrt{3}\right)}}{\sqrt{5}-\sqrt{3}}=}$${\displaystyle \frac{2(\sqrt{5}+\sqrt{3})}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}=}$${\displaystyle \frac{2(\sqrt{5}+\sqrt{3})}{2}=}$${\displaystyle \sqrt{5}+\sqrt{3}}$;

4) ${\displaystyle \frac{3^{\mathrm{\setminus }\left(\sqrt{8}-\sqrt{5}\right)}}{\sqrt{8}+\sqrt{5}}=}$${\displaystyle \frac{3(\sqrt{8}-\sqrt{5})}{{\left(\sqrt{8}\right)}^2-{\left(\sqrt{5}\right)}^2}=}$${\displaystyle \frac{3(\sqrt{8}-\sqrt{5})}{3}=}$${\displaystyle \sqrt{8}-\sqrt{5}=}$${\displaystyle 2\sqrt{2}-\sqrt{5}}$;

5) ${\displaystyle \frac{1^{\mathrm{\setminus }\left(1-\sqrt{m}\right)}}{1+\sqrt{m}}=}$${\displaystyle \frac{1-\sqrt{m}}{1^2-{\left(\sqrt{m}\right)}^2}=}$${\displaystyle \frac{1-\sqrt{m}}{1-m}}$;

6) ${\displaystyle \frac{1^{\mathrm{\setminus }\left(\sqrt{a}+\sqrt{c}\right)}}{\sqrt{a}-\sqrt{c}}=}$${\displaystyle \frac{\sqrt{a}+\sqrt{c}}{{\left(\sqrt{a}\right)}^2-{\left(\sqrt{c}\right)}^2}=}$${\displaystyle \frac{\sqrt{a}+\sqrt{c}}{a-c}}$.