Question

Только левый столбик

Answer 1

$1)\; \; (x-1)(x+9)<0\qquad +++(-9)---(1)+++\\\\x\in (-9,1)\\\\2)\; \; x^2>9\; ,\; \; (x-3)(x+3)>0\; ,\\\\+++(-3)---(3)+++\qquad x\in (-\infty ,-3)\cup (3,+\infty )\\\\3)\; \; \frac{x(x-1)}{3-x}\leq 0\; ,\; \; \frac{x(x-1)}{-(x-3)}\leq 0\; ,\; \; \frac{x(x-1)}{x-3}\geq 0\\\\---[\, 0\, ]+++[\, 1\, ]---(3)+++\qquad x\in [\, 0,1\, ]\cup (3,+\infty )\\\\4)\; \; x^2(5x-4)(x+7)<0\; ,\\\\+++(-7)---(0)---(4/5)+++\\\\x\in (-7,0)\cup (0,\frac{4}{5})$

$5)\; \; y=\sqrt{\frac{-3x+x^2-4}{9-x^2}}\; \; ,\; \; OOF:\; \; \left \{ {{\frac{-3x+x^2-4}{9-x^2}\geq 0} \atop {9-x^2\ne 0}} \right. \\\\x^2-3x-4=0\; \; \to \; \; x_1=-1\; ,\; x_2=4\; \; (teorema\; Vieta)\\\\9-x^2\ne 0\; \to \; \; x^2\ne 9\; ,\; \; x\ne \pm 3\\\\\frac{x^2-3x-4}{-(x^2-9)}\geq 0\; \; ,\; \; \frac{(x+1)(x-4)}{(x-3)(x+3)}\leq 0\; \; ,\\\\+++(-3)---[\, -1\, ]+++(3)---[\, 4\, ]+++\\\\x\in (-3,-1\, ]\cup (3,4\, ]$