Тез арада керек болып тур өтінем алгебрадан

Решите задачу:

$3)\; \; \frac{x^2-6,5x+13}{x^2-0,5x-14}\ \textless \ 1\\\\\frac{x^2-6,5x+13}{x^2-0,5x-14}-1\ \textless \ 0\\\\\frac{x^2-6,5x+13-x^2+0,5x+14}{x^2-0,5x-14}\ \textless \ 0\\\\\frac{-6x+27}{(x+3,5)(x-4)}\ \textless \ 0\; \; ,\; \; \frac{-6(x-4,5)}{(x+3,5)(x-4)}\ \textless \ 0\\\\+++(-3,5)---(4)+++(4,5)---\\\\x\in (-3,5\, ;\, 4)\cup (4,5\, ;+\infty )$

$4)\; \; (x+9)(x-11)^2(x-12)\ \textless \ 0\\\\znaki:\; \; \; +++(-9)---(11)---(12)+++\\\\x\in (-9,11)\cup (11,12)$

$5)\; \; y= \frac{\sqrt{x^3-9x}}{x-5}= \frac{\sqrt{x(x-3)(x+3)}}{x-5}\\\\\\\left \{ {{x(x-3)(x+3) \geq 0} \atop {x-5\ne 0}} \right. \; \left \{ {{x\in [-3.0\, ]\cup [\, 3,+\infty )} \atop {x\ne 5}} \right. \; \; \to \; \; x\in [-3,0\, ]\cup [\, 3,5)\cup (5,+\infty )\\\\\\x(x-3)(x+3) \geq 0\; \; \; ---(-3)+++(0)---(3)+++\\\\x\in [-3,0\, ]\cup [\, 3,+\infty )$