Question

А)-х^2-4< 4х б)х^2-2х> -1 в)6х^2> 15-х

Answer 1

$1)\; \; -x^2-4\ \textless \ 4x\\\\x^2+4x+4\ \textgreater \ 0\\\\(x+2)^2\ \textgreater \ 0\; \; \Rightarrow\; \; x+2\ne 0\; \; \Rightarrow \; \; x\nw -2\\\\x\in (-\infty ,-2)\cup (-2,+\infty )\\\\2)\; \; x^2-2x\ \textgreater \ -1\\\\x^2-2x+1\ \textgreater \ 0\\\\(x-1)^2\ \textgreater \ 0\; \; \Rightarrow \; \; x\ne 1\\\\x\in (-\infty ,1)\cup (1,+\infty )$

$3)\; \; 6x^2\ \textgreater \ 15-x\\\\6x^2+x-15\ \textgreater \ 0\\\\\sqrt{D}=19\; ,\; x_1=\frac{-1-19}{12}=-\frac{5}{3}\; ,\; x_2=\frac{3}{2}\\\\6(x+\frac{5}{3})(x-\frac{3}{2})\ \textgreater \ 0\\\\+++(-\frac{5}{3})---(\frac{3}{2})+++\\\\x\in (-\infty ,-\frac{5}{3})\cup (\frac{3}{2},+\infty )$