33 Вправа 33 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 320

33. Спростіть вираз:

1) ${\displaystyle \frac{\sqrt{12}-\sqrt{27}}{\sqrt{15}-\sqrt{3}}=}$${\displaystyle \frac{\sqrt{3}\cdot \sqrt{4}-\sqrt{3}\cdot \sqrt{9}}{\sqrt{3}(\sqrt{5}-1)}=}$${\displaystyle \frac{\cancel{\sqrt{3}}(2-3)}{\cancel{\sqrt{3}}(\sqrt{5}-1)}=}$${\displaystyle \frac{-1}{\sqrt{5}-1}}$;

2) ${\displaystyle \frac{\sqrt{28}+2}{\sqrt{7}+1}=}$${\displaystyle \frac{\sqrt{4\cdot 7}+2}{\sqrt{7}+1}=}$${\displaystyle \frac{2\sqrt{7}+2}{\sqrt{7}+1}=}$${\displaystyle \frac{2\cancel{\left(\sqrt{7}+1\right)}}{\cancel{\sqrt{7}+1}}=}$${\displaystyle 2}$;

3) ${\displaystyle \frac{{\sqrt{3}}^{\mathrm{\setminus }\sqrt{3}-\sqrt{2}}}{\sqrt{3}+\sqrt{2}}-\frac{{\sqrt{2}}^{\mathrm{\setminus }\sqrt{3}+\sqrt{2}}}{\sqrt{3}-\sqrt{2}}=}$${\displaystyle \frac{\sqrt{3}(\sqrt{3}-\sqrt{2})-\sqrt{2}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}=}$${\displaystyle \frac{3-\sqrt{6}-\sqrt{6}-2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=}$${\displaystyle \frac{1-2\sqrt{6}}{3-2}=}$${\displaystyle 1-2\sqrt{6}}$;

4) ${\displaystyle \frac{{\sqrt{7}+2}^{\mathrm{\setminus }\sqrt{7}+2}}{\sqrt{7}-2}-\frac{{\sqrt{7}-2}^{\mathrm{\setminus }\sqrt{7}-2}}{\sqrt{7}+2}=}$${\displaystyle \frac{{(\sqrt{7}+2)}^2-{(\sqrt{7}-2)}^2}{(\sqrt{7}-2)(\sqrt{7}+2)}=}$${\displaystyle \frac{\cancel{7}+4\sqrt{7}\cancel{+4}-\left(\cancel{7}-4\sqrt{7}\cancel{+4}\right)}{{\left(\sqrt{7}\right)}^2-2^2}=}$${\displaystyle \frac{8\sqrt{7}}{7-4}=}$${\displaystyle \frac{8\sqrt{7}}{3}}$.