![1)\; \; \frac{(x^2-7x-8)(x-\sqrt{63})^3}{(x+2)^2(4-x)} \geq 0\\\\ \frac{(x-8)(x+1)(x-\sqrt{63})^3}{(x+2)^2(x-4)} \leq 0\; ,\; \; x\ne -2\; ,\; \; x\ne 4\\\\+++(-2)+++[-1\, ]---(4)+++[\sqrt{63}]---[\, 8\, ]+++\\\\x\in [-1,4)\cup [\sqrt{63},8\, ]](https://tex.z-dn.net/?f=1%29%5C%3B+%5C%3B++%5Cfrac%7B%28x%5E2-7x-8%29%28x-%5Csqrt%7B63%7D%29%5E3%7D%7B%28x%2B2%29%5E2%284-x%29%7D++%5Cgeq+0%5C%5C%5C%5C+%5Cfrac%7B%28x-8%29%28x%2B1%29%28x-%5Csqrt%7B63%7D%29%5E3%7D%7B%28x%2B2%29%5E2%28x-4%29%7D+%5Cleq+0%5C%3B+%2C%5C%3B+%5C%3B+x%5Cne+-2%5C%3B+%2C%5C%3B+%5C%3B+x%5Cne+4%5C%5C%5C%5C%2B%2B%2B%28-2%29%2B%2B%2B%5B-1%5C%2C+%5D---%284%29%2B%2B%2B%5B%5Csqrt%7B63%7D%5D---%5B%5C%2C+8%5C%2C+%5D%2B%2B%2B%5C%5C%5C%5Cx%5Cin+%5B-1%2C4%29%5Ccup+%5B%5Csqrt%7B63%7D%2C8%5C%2C+%5D "1)\; \; \frac{(x^2-7x-8)(x-\sqrt{63})^3}{(x+2)^2(4-x)} \geq 0\\\\ \frac{(x-8)(x+1)(x-\sqrt{63})^3}{(x+2)^2(x-4)} \leq 0\; ,\; \; x\ne -2\; ,\; \; x\ne 4\\\\+++(-2)+++[-1\, ]---(4)+++[\sqrt{63}]---[\, 8\, ]+++\\\\x\in [-1,4)\cup [\sqrt{63},8\, ]")
0\\\\|x|=5\; \; \to \; \; x=\pm 5" alt="2)\; \; x^2-4|x|-2= \frac{15}{x^2-4|x|} \\\\ODZ:\; x^2-4|x|\ne 0\; ,\; \; |x|^2-4|x|\ne 0\; ,\; \; |x|\cdot (|x|-4)\ne 0\; ,\\\\|x|\ne0\; ,|x|\ne 4\; \; \to \; \; x\ne 0,\; x\ne \pm 4\\\\Tak\; kak\; \; x^2=|x|^2\; ,\; to\; \; |x|^2-4|x|-2= \frac{15}{|x|^2-4|x|} \\\\oboznachim\; \; a=|x|^2-4|x|\; \; \to \; \; a-2= \frac{15}{a} \; ,\; a\ne 0\; \; \to \\\\a^2-2a-15=0\; \; \to \; \; a_1=5\; ,\; a_2=-3\; \; (teorema\; Vieta)\\\\a)\; \; |x|^2-4|x|=5\; ,\; \; |x|^2-4|x|-5=0\\\\t=|x|\geq 0\; ,\; \; t^2-4t-5=0\; ,\; t_1=-1<0\; ,\; \; t_2=5>0\\\\|x|=5\; \; \to \; \; x=\pm 5" align="absmiddle" class="latex-formula">
