Question

Решите пожалуйста второй и третий номер.

Answer 1

$1); ; sqrt{0,75}-sqrt{108}-frac{1}{32}sqrt{192}+sqrt{147}=\\=sqrt{3cdot 0,25}-sqrt{3cdot 36}-frac{1}{32}cdot sqrt{3cdot 64}+sqrt{3cdot 49}=\\=0,5sqrt3-6sqrt3-frac{8}{32}sqrt3+7sqrt3=sqrt3cdot (0,5-6-0,25+7)=1,25cdot sqrt3\\2); ; (sqrt7-3)^2(16+6sqrt7)-4sqrt{3frac{1}{16}}=(16-6sqrt7)(16+6sqrt7)-4sqrt{frac{49}{16}}=\\=16^2-36cdot 7-4cdot frac{7}{4}=256-252-7=-3$

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

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

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

Число х=-4 удовлетворяет заданному неравенству.