Question

помогите пожалуйста!!!!

Answer 1

$3)\; \; x^2-x-6<0\; \; \to \; \; \; (x+2)(x-3)<0\; \; ,\\\\+++(-2)---(3)+++\ \ \ \ \ \ \underline {\ x\in (-2;3\, )\ }$

$4)\ \ \ \left\{\begin{array}{l}3-x>1\\x^2-9\leq 7\end{array}\right\; \ \left\{\begin{array}{l}3-1>x\\x^2-16\leq 0\end{array}\right\ \ \left\{\begin{array}{l}2>x\\(x-4)(x+4)\leq 0\end{array}\right\\\\\left\{\begin{array}{l}x<2\\x\in [-4\, ;\, 4\; ]\end{array}\right\ \ \ \Rightarrow \ \ \ \ \underline {\ x\in [-4\, ;2\, )\; }$

$5)\ \ \ \dfrac{x^2-x}{(x-5)(x+3)}\leq 0\ \ ,\ \ \ \ \dfrac{x(x-1)}{(x-5)(x+3)}\leq 0\ \ ,\ \ \ ODZ:\ x\ne 5,\; x\ne -3\ ,\\\\\\x_1=0\ ,\ x_2=1\ ,\ x_3=5\ ,\ x_4=-3\\\\znaki:\ \ \ +++(-3)---[\; 0\; ]+++[\; 1\; ]---(5)+++\\\\\underline {\; x\in (-3\, ;\; 0\; ]\cup [\; 1\, ;\, 5\, )\; }$