math-public:metod_intervalov

Четная точка

Номер	Условие	Ответ
1.	$5x^2>x^4$	$(-\sqrt{5}; 0)\cup(0, \sqrt{5})$
2.	$\dfrac{5x^2-4x}{(11-2x)^2}\leqslant0$	$\left[0; \dfrac{4}{5}\right]$
3.	$(5-x)(x^2-11x+30)^2\leqslant0$	$[5; +\infty)$
4.	$(x^2-12x+32)(x^2-24x+144)>0$	$(-\infty; 4)\cup(8; 12)\cup(12; +\infty)$
5.	$\dfrac{(x^2-17x+66)^2(6-x)}{(3x+2)^2(9x^2-4)}<0$	$\left(-\dfrac{2}{3}; \dfrac{2}{3}\right)\cup(6; 11)\cup(11; +\infty)$
6.	$\dfrac{x+14}{x-2}+\dfrac{x-4}{x}<0$	$(0; 2)$
7.	$\dfrac{x+6}{x+3}-\dfrac{x+5}{3-x}\leqslant\dfrac{x^2+9x-4}{x^2-9}$	$(-3; 3)$
8.	$\dfrac{12}{(x^2-7x+10)(x-1)}>\dfrac{2x-8}{x-1}-\dfrac{x-1}{x-2}-\dfrac{x-3}{x-5}$	$(1; 2)\cup(5; +\infty)$
9.	$\dfrac{1}{x+4}\leqslant\dfrac{2}{4x+15}+\dfrac{1}{x+6}$	$(-6; -4)\cup\left(-\dfrac{15}{4}; +\infty\right)$
10.	$\dfrac{43 x+74}{x^2-4}+\dfrac{17-4 x}{x-2}<\dfrac{52-4x}{x-1}$	$(-\infty; -5)\cup(-5; -2)\cup(1; 2)$
11.	$\dfrac{5}{x-3}-\dfrac{x^2+18}{x^3-27}<\dfrac{x}{x^2+3x+9}$	$(-\infty; -3)\cup(-3; 3)$
12.	$\dfrac{(x^2+4x)^2-(4x^2-x)^2}{(25x^2+30x+9)^2}>0$	$\left(-\dfrac{3}{5}; 0\right)\cup\left(0; \dfrac{5}{3}\right)$
13.	$\left(\dfrac{x-5}{x+3}\right)^2>\left(\dfrac{x-5}{x-8}\right)^2$	$(-\infty; -3)\cup\left(-3; \dfrac{5}{2}\right)$
14.	$\dfrac{(x^2-16x+48)(x^2-14x+24)}{(x+5)^3(x^2-25)}<0$	$(-\infty; -5)\cup(-5; 2)\cup(4; 5)$
15.	$\dfrac{19 x+44}{(x^2+4)}-\dfrac{25}{2 (x-2)}+\dfrac{3}{2 (x+2)}>0$	$(-2; 2)\cup(4; +\infty)$
16.	$\dfrac{x+3}{x}\cdot\left(\dfrac{10 (2 x-1)}{x^2+1}+x-\dfrac{12}{x}+6\right)\leqslant0$	$(-\infty; 0)\cup(0; 1]$
17.	$\dfrac{2(7x-52)}{x^2+4}+16\leqslant\dfrac{7}{2-x}-\dfrac{27}{x+2}$	$[-5; -2)\cup[0; 2)$
18.	$\dfrac{x+4}{x-3}-\dfrac{1}{x^2+3x+9}>\dfrac{x^2+11x+39}{x^3-27}$	$(-\infty; -3)\cup(-3; 0)\cup(3; +\infty)$
19.	$4x^2+24x+\dfrac{625}{x-5}\geqslant\dfrac{1}{x-1}-124$	$(-\infty; 1)\cup(5; +\infty)$
20.	$\dfrac{2}{2x^2+x-3}-\dfrac{1}{9x^2+6x+1}\geqslant\dfrac{3}{(3x+1)(1-x)}$	$\left(-\infty; -\dfrac{3}{2}\right)\cup\left[\dfrac{-11-\sqrt{2}}{17}; \dfrac{\sqrt{2}-11}{17}\right]\cup(1; +\infty)$