8. Найдите значение выражения 1 \frac{(7\sqrt{11})^2}{110}; 2 \frac{48}{(2\sqrt{6})^2}; 3 (\sqrt{23}-4)(\sqrt{23}+4); 4 (\sqrt{15}-\sqrt{7})(\sqrt{15}+\sqrt{7}); 5 (√14-3)2+6√14 6\frac{1}{4+\sqrt{14}}+\frac{1}{4-\sqrt{14}}; 7\frac{1}{\sqrt{37}-6}-\frac{1}{\sqrt{37}+6}.

Ответ: 4.9

Разбираемся:

1. \[\frac{(7\sqrt{11})^2}{110} = \frac{49 \cdot 11}{110} = \frac{49}{10} = 4.9\]
2. \[\frac{48}{(2\sqrt{6})^2} = \frac{48}{4 \cdot 6} = \frac{48}{24} = 2\]
3. \[(\sqrt{23}-4)(\sqrt{23}+4) = (\sqrt{23})^2 - 4^2 = 23 - 16 = 7\]
4. \[(\sqrt{15}-\sqrt{7})(\sqrt{15}+\sqrt{7}) = (\sqrt{15})^2 - (\sqrt{7})^2 = 15 - 7 = 8\]
5. \[(\sqrt{14}-3)^2 + 6\sqrt{14} = 14 - 6\sqrt{14} + 9 + 6\sqrt{14} = 14 + 9 = 23\]
6. \[\frac{1}{4+\sqrt{14}} + \frac{1}{4-\sqrt{14}} = \frac{(4-\sqrt{14}) + (4+\sqrt{14})}{(4+\sqrt{14})(4-\sqrt{14})} = \frac{8}{16 - 14} = \frac{8}{2} = 4\]
7. \[\frac{1}{\sqrt{37}-6} - \frac{1}{\sqrt{37}+6} = \frac{(\sqrt{37}+6) - (\sqrt{37}-6)}{(\sqrt{37}-6)(\sqrt{37}+6)} = \frac{12}{37 - 36} = \frac{12}{1} = 12\]

Ответ: 4.9, 2, 7, 8, 23, 4, 12