Решите пожалуйста второй и третий номер.

Question 1

Question 2

$1)\; \; \sqrt{0,75}-\sqrt{108}-\frac{1}{32}\sqrt{192}+\sqrt{147}=\\\\=\sqrt{3\cdot 0,25}-\sqrt{3\cdot 36}-\frac{1}{32}\cdot \sqrt{3\cdot 64}+\sqrt{3\cdot 49}=\\\\=0,5\sqrt3-6\sqrt3-\frac{8}{32}\sqrt3+7\sqrt3=\sqrt3\cdot (0,5-6-0,25+7)=1,25\cdot \sqrt3\\\\2)\; \; (\sqrt7-3)^2(16+6\sqrt7)-4\sqrt{3\frac{1}{16}}=(16-6\sqrt7)(16+6\sqrt7)-4\sqrt{\frac{49}{16}}=\\\\=16^2-36\cdot 7-4\cdot \frac{7}{4}=256-252-7=-3$

$3)\; \; \frac{1-\sqrt{21}}{\sqrt3+\sqrt7} + \frac{26}{3\sqrt3-1} -\sqrt{21}-(\sqrt7-1)(1-\sqrt3)=\\\\= \frac{(1-\sqrt{21})(\sqrt7-\sqrt3)}{(\sqrt7+\sqrt3)(\sqrt7-\sqrt3)} + \frac{26(3\sqrt3+1)}{(3\sqrt3-1)(3\sqrt3+1)} -\sqrt{21}-(\sqrt7-\sqrt{21}-1+\sqrt3)=\\\\= \frac{\sqrt7-\sqrt3-7\sqrt3+3\sqrt7}{7-3} + \frac{26(3\sqrt3+1)}{27-1} -\sqrt{21}-\sqrt7+\sqrt{21}+1-\sqrt3=\\\\= \frac{4\sqrt7-8\sqrt3}{4} +3\sqrt3+1-\sqrt7+1-\sqrt3=\\\\=\sqrt7-2\sqrt3+2\sqrt3+2-\sqrt7=2$

$4)\; \; \sqrt{10-4\sqrt6}-\sqrt{10+4\sqrt6}=\sqrt{(2-\sqrt6)^2}-\sqrt{(2+\sqrt6)^2}=\\\\=|2-\sqrt6|-|2+\sqrt6|=(\sqrt6-2)-(2+\sqrt6)=-4\\\\\\5x^2+34x+51\ \textless \ 0\\\\D/4=17^2-5\cdot 51=-255=34\\\\x_1=\frac{-17-\sqrt{34}}{5}\approx -4,57\; ;\; \; x_2=\frac{-17+\sqrt{34}}{5}\approx -2,23\\\\x\in (\frac{-17-\sqrt{34}}{5}\; ;\; \frac{-17+\sqrt{34}}{5})$

$-4\in (\frac{-17-\sqrt{34}}{5}\; ;\; \frac{-17+\sqrt{34}}{5})$

Число х=-4 удовлетворяет заданному неравенству.