663 Вправа 663Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 195

663. Звільніться від ірраціональності в знаменнику дробу:

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

2) ${\displaystyle \frac{7^{\mathrm{\setminus }\left(\sqrt{2}-3\right)}}{\sqrt{2}+3}=}$${\displaystyle \frac{7(\sqrt{2}-3)}{{\left(\sqrt{2}\right)}^2-3^2}=}$${\displaystyle \frac{7(\sqrt{2}-3)}{2-9}=}$${\displaystyle -(\sqrt{2}-3)=}$${\displaystyle 3-\sqrt{2}}$;

3) ${\displaystyle \frac{4^{\mathrm{\setminus }\left(\sqrt{7}+\sqrt{5}\right)}}{\sqrt{7}-\sqrt{5}}=}$${\displaystyle \frac{4(\sqrt{7}+\sqrt{5})}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{5}\right)}^2}=}$${\displaystyle \frac{4(\sqrt{7}+\sqrt{5})}{7-5}=}$${\displaystyle 2\sqrt{7}+2\sqrt{5}}$;

4) ${\displaystyle \frac{7^{\mathrm{\setminus }\left(\sqrt{10}-\sqrt{3}\right)}}{\sqrt{10}+\sqrt{3}}=}$${\displaystyle \frac{7(\sqrt{10}-\sqrt{3})}{{\left(\sqrt{10}\right)}^2-{\left(\sqrt{3}\right)}^2}=}$${\displaystyle \frac{7(\sqrt{10}-\sqrt{3})}{10-3}=}$${\displaystyle \sqrt{10}-\sqrt{3}}$;

5) ${\displaystyle \frac{2^{\mathrm{\setminus }\left(\sqrt{11}+\sqrt{2}\right)}}{\sqrt{11}-\sqrt{2}}=}$${\displaystyle \frac{2(\sqrt{11}+\sqrt{2})}{{\left(\sqrt{11}\right)}^2-{\left(\sqrt{2}\right)}^2}=}$${\displaystyle \frac{2(\sqrt{11}+\sqrt{2})}{11-2}=}$${\displaystyle \frac{2\left(\sqrt{11}+\sqrt{2}\right)}{9}}$;

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

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

8) ${\displaystyle \frac{1^{\mathrm{\setminus }\left(\sqrt{y}-\sqrt{x}\right)}}{\sqrt{y}+\sqrt{x}}=}$${\displaystyle \frac{\sqrt{y}-\sqrt{x}}{{\left(\sqrt{y}\right)}^2-{\left(\sqrt{x}\right)}^2}=}$${\displaystyle \frac{\sqrt{y}-\sqrt{x}}{y-x}}$.