Решите неравенство: X^2 + 2√x^2-6x ‹= 6x+24 √ - корень икс в квадрате минус шесть икс ‹=...

Ответ:

$x^2+2\sqrt{x^2-6x}\leq 6x+24\ \ ,\\\\ODZ:\ x^2-6x\geq 0\ ,\ x(x-6)\geq 0\ \ ,\ \ x\in (-\infty ;\, 0\, ]\cup [\, 6\, ;+\infty)\\\\(x^2-6x)+2\sqrt{x^2-6x}-24\leq 0\\\\t=\sqrt{x^2-6x}\geq 0\ \ ,\ \ \ t^2+2t-24\leq 0\ ,\ \ D/4=1+24=25\ ,\ t_1=-6\ ,\ t_2=4\\\\(t+6)(t-4)\leq 0\ \ \ \ \ \ \ +++[-6\, ]---[\, 4\, ]+++$

$t\in [\ -6\, ;\, 4\ ]\ ,\ t\geq 0\ \ \Rightarrow \ \ \ \left\{\begin{array}{ccc}x^2-6x\geq 0\\x^2-6x\leq 4\end{array}\right\ \ \left\{\begin{array}{l}x(x-6)\geq 0\\x^2-6x-4\leq 0\end{array}\right$

$x(x-6)\geq 0\ \ ,\ \ x\in (-\infty ;\ 0\ ]\cup [\ 6\, ;+\infty )\\\\x^2-6x-4=0\ \ ,\ \ D/4=13\ ,\ x_1=3-\sqrt{13}\approx -0,61\ ,\ x_2=3+\sqrt{13}\approx 6,61\\\\\left\{\begin{array}{l}x\in (-\infty ;\ 0\, ]\cup [\ 6\ ;+\infty )\\x\in [\, 3-\sqrt{13}\, ;\, 3+\sqrt{13}\, \, ]\end{array}\right$

$x\in [\, 3-\sqrt{13}\, ;\ 0\ ]\cup [\ 6\, ;3+\sqrt{13}\ ]$