8. Найдите значение выражения 1 (7-11)2 110 ; 3 (√23-4)(√23+4); 6 4+14 1 + 4-14 1 ; 2 48 (2/6)2 ; 4 (15-17)(15+√7); 7 √37-6 1 - √37+6 1 ; 5 (/14-3)2+6/14

1. Вычислим значение выражения

1) $$\frac{(7\sqrt{11})^2}{110} = \frac{7^2 \cdot (\sqrt{11})^2}{110} = \frac{49 \cdot 11}{110} = \frac{49 \cdot 11}{10 \cdot 11} = \frac{49}{10} = 4.9$$

2) $$\frac{48}{(2\sqrt{6})^2} = \frac{48}{2^2 \cdot (\sqrt{6})^2} = \frac{48}{4 \cdot 6} = \frac{48}{24} = 2$$

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) $$(\sqrt{14}-3)^2 + 6\sqrt{14} = (\sqrt{14})^2 - 2 \cdot 3 \cdot \sqrt{14} + 3^2 + 6\sqrt{14} = 14 - 6\sqrt{14} + 9 + 6\sqrt{14} = 14 + 9 = 23$$

6) $$\frac{1}{4+\sqrt{14}} + \frac{1}{4-\sqrt{14}} = \frac{4-\sqrt{14} + 4 + \sqrt{14}}{(4+\sqrt{14})(4-\sqrt{14})} = \frac{8}{4^2 - (\sqrt{14})^2} = \frac{8}{16-14} = \frac{8}{2} = 4$$

7) $$\frac{1}{\sqrt{37}-6} - \frac{1}{\sqrt{37}+6} = \frac{\sqrt{37}+6 - (\sqrt{37}-6)}{(\sqrt{37}-6)(\sqrt{37}+6)} = \frac{\sqrt{37}+6 - \sqrt{37}+6}{(\sqrt{37})^2 - 6^2} = \frac{12}{37-36} = \frac{12}{1} = 12$$

Ответ: 1) 4.9; 2) 2; 3) 7; 4) 8; 5) 23; 6) 4; 7) 12