Question

СРОЧНО!!!! ДАМ 30 БАЛЛОВ!!!​

Answer 1

$1)\; \; y\leq 0\; \; ,\; \; \sqrt{7y^2}=|y|\cdot \sqrt7=-y\, \sqrt7\\\\2)\; \; \sqrt{32a^8}=\sqrt{2^4\cdot 2\cdot (a^4)^2}=2^2\cdot |a^4|\cdot \sqrt2=4\, a^4\sqrt2\\\\3)\; \; \sqrt{-b^{15}}=\sqrt{-(b^7)^2\cdot b}=|b^7|\cdot \sqrt{-b}}=-b^7\cdot \sqrt{-b}\\\\\Big(-b\geq 0\; \to \; b\leq0\; \; \to \; \; |b|=-b\; \; \to \; \; |b^7|=-b^7\; \Big)$

$4)\; \; x>0\; \; \; ,\; \; \sqrt{-x^{14}y^3}=\sqrt{-(x^7)^2\cdot y^2\cdot y}=|x^7|\cdot |y|\cdot \sqrt{-y}=\Big[\; -y\geq 0\; \to \\\\y\leq 0\; \; \to \; \; |y|=-y\; \; ;\; \; \; x>0\; \to \; x^7>0\; \to \; \; |x^7|=x^7\; \Big]=-x^7y\sqrt{-y}\\\\\\5)\; \; \sqrt{(5-\sqrt{12})^2}+\sqrt{(3-\sqrt{12})^2}=|\underbrace {5-\sqrt{12}}_{>0}|+|\underbrace {3-\sqrt{12}}_{<0}|=\\\\=(5-\sqrt{12})+(-3+\sqrt{12})=5-3=2$