569 Вправа 569Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 164

569. Спростіть вираз:

1) ${\displaystyle \frac{\sqrt{45}-\sqrt{4,5}}{\sqrt{10}-1}=}$${\displaystyle \frac{\sqrt{4,5}(\sqrt{10}-1)}{\sqrt{10}-1}=}$${\displaystyle \sqrt{4,5}=}$${\displaystyle 1,5\sqrt{2}}$;

2) ${\displaystyle \frac{\sqrt{48}+2\sqrt{2}}{\sqrt{6}+1}=}$${\displaystyle \frac{2\sqrt{2}\cdot \sqrt{6}+2\sqrt{2}}{\sqrt{6}+1}=}$${\displaystyle \frac{2\sqrt{2}(\sqrt{6}+1)}{\sqrt{6}+1}=}$${\displaystyle 2\sqrt{2}}$;

3) ${\displaystyle \frac{\sqrt{3}-\sqrt{27}}{\sqrt{0,12}+\sqrt{3}}=}$${\displaystyle \frac{\sqrt{3}-3\sqrt{3}}{0,2\sqrt{3}+\sqrt{3}}=}$${\displaystyle \frac{-2\sqrt{3}}{1,2\sqrt{3}}=}$${\displaystyle -\frac{5}{3}=}$${\displaystyle -1\frac{2}{3}}$;

4) ${\displaystyle \frac{\sqrt{225}+2\sqrt{625}}{\sqrt{169}}=}$${\displaystyle \frac{15+2\cdot 25}{13}=}$${\displaystyle \frac{65}{13}=}$${\displaystyle 5}$;

5) ${\displaystyle \frac{\sqrt{0,3}-\sqrt{0,6}}{1-\sqrt{2}}=}$${\displaystyle \frac{\sqrt{0,3}(1-\sqrt{2})}{1-\sqrt{2}}=}$${\displaystyle \sqrt{0,3}}$;

6) ${\displaystyle \frac{\sqrt{4,41}-2\sqrt{2,1}+1}{1-\sqrt{2,1}}=}$${\displaystyle \frac{{\left(\sqrt{2,1}\right)}^2-2\sqrt{2,1}+1}{1-\sqrt{2,1}}=}$${\displaystyle \frac{{(1-\sqrt{2,1})}^2}{1-\sqrt{2,1}}=}$${\displaystyle 1-\sqrt{2,1}}$;

7) ${\displaystyle \sqrt{1,44}=}$${\displaystyle {\left(\sqrt{1,2}\right)}^2}$, ${\displaystyle 2\sqrt{3,6}=}$${\displaystyle 2\sqrt{1,2}\cdot \sqrt{3}}$, ${\displaystyle 2\sqrt{0,3}=}$${\displaystyle \sqrt{1,2}}$;

${\displaystyle \frac{\sqrt{1,44}+2\sqrt{3,6}+3}{\sqrt{3}+2\sqrt{0,3}}-0,4\sqrt{7,5}=}$${\displaystyle \frac{{(\sqrt{1,2}+\sqrt{3})}^2}{\sqrt{3}+\sqrt{1,2}}-\sqrt{0,16\cdot 7,5}=}$${\displaystyle \sqrt{1,2}+\sqrt{3}-\sqrt{1,2}=}$${\displaystyle \sqrt{3}}$;

8) ${\displaystyle \frac{2}{5+2\sqrt{6}}+\frac{2}{5-2\sqrt{6}}=}$${\displaystyle \frac{2(5-2\sqrt{6})+2(5+2\sqrt{6})}{25-24}=}$${\displaystyle \frac{20}{1}=}$${\displaystyle 20}$;

9) ${\displaystyle \frac{\sqrt{5}+1}{\sqrt{15}}-\frac{1}{\sqrt{3}}=}$${\displaystyle \frac{\sqrt{5}+1}{\sqrt{15}}-\frac{\sqrt{5}}{\sqrt{15}}=}$${\displaystyle \frac{\sqrt{5}+1-\sqrt{5}}{\sqrt{15}}=}$${\displaystyle \frac{1}{\sqrt{15}}=}$${\displaystyle \frac{\sqrt{15}}{15}}$;

10) ${\displaystyle \frac{1}{8+2\sqrt{7}}+\frac{1}{8-2\sqrt{7}}=}$${\displaystyle \frac{8-2\sqrt{7}+8+2\sqrt{7}}{64-28}=}$${\displaystyle \frac{16}{36}=}$${\displaystyle \frac{4}{9}}$.