Решить неравенство: a) x+9>8-4x b) x-8<3x+5 решить уравнение: a) x^2+4x-45=0 ...

Решите задачу:

$1)\; \; x+9\ \textgreater \ 8-4x\\5x\ \textgreater \ -1\\x\ \textgreater \ -0,2\; \; ,\quad x\in (-0,2\; ;\; +\infty )\\\\1b)\; \; x-8\ \textless \ 3x+5\\-13\ \textless \ 2x\\x\ \textless \ -6,5\\\\2a)\; \; x^2+4x-45=0\\x_1=-9,\; \; x_2=5\; \; \; (teorema\; Vieta)\\\\2b)\; \; x^2-9x-52=0\\D=81+4\cdot 52=289=17^2\; ,\; \; x_1=\frac{9-17}{2}=-4\; ,\; x_2=\frac{9+17}{2}=13$

$3)\; \; \frac{5x-5\sqrt3}{3-x^2}=\frac{5(x-\sqrt3)}{(\sqrt3-x)(\sqrt3+x)}=-\frac{5}{\sqrt3+x}\\\\\frac{\sqrt{15}-5}{\sqrt6-\sqrt{10}}=\frac{\sqrt5(\sqrt3-\sqrt5)}{\sqrt2(\sqrt3-\sqrt5)}=\sqrt{\frac{5}{2}}=\sqrt{2,5}\\\\4a)\; \; 7=\sqrt{49}\\\\49\ \textgreater \ 48\; \; \to \; \; \sqrt{49}\ \textgreater \ \sqrt{48}\; \; \; \to \; \; \; 7\ \textgreater \ \sqrt{48}\\\\4b)\; \; \sqrt{81}=9\\\\9\ \textgreater \ 8\; \; \; \to \; \; \; \sqrt{81}\ \textgreater \ 8\\\\5)\; \; 44,301=4,4301\cdot 10\\\\0,483=4,83\cdot 10^{-1}$