Question

Решите неравенство:...........................

Answer 1

Ответ:

$\dfrac{2\, log_4(x^2-3x)}{log_4(x-5)^2}\leq 1\ \ \ ,\ \ \ ODZ:\ \left\{\begin{array}{l}x^2-3x>0\\(x-5)^2>0\\log+4(x-5)^2\ne 0\end{array}\right\\\\\\\left\{\begin{array}{l}x(x-3)>0\\(x-5)^2\ne 0\\(x-5)^2\ne 1\end{array}\right\ \ \left\{\begin{array}{l}x(x-3)>0\\(x-5)^2\ne 0\\(x-5)^2\ne 1\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;0)\cup (3\, ;+\infty )\\x\ne 5\\x\ne 4\ ,\ x\ne 6\end{array}\right$

$\dfrac{2\, log_4(x^2-3x)-log_4(x-5)^2}{log+4(x-5)^2}\leq 0\ \ ,\ \ \ \dfrac{log_4\dfrac{(x^2-3x)^2}{(x-5)^2}}{log_4(x-5)^2}\leq 0\\\\\\a)\\\left\{\begin{array}{l}log_4\dfrac{(x^2-3x)^2}{(x-5)^2}\leq 0\\log_4(x-5)^2>0\end{array}\right\ \left\{\begin{array}{l}\Big(\dfrac{x^2-3x}{x-5}\Big)^2\leq 1\\(x-5)^2>1\end{array}\right\ \left\{\begin{array}{l}\Big(\dfrac{x^2-3x}{x-5}-1\Big)\Big(\dfrac{x^2-3x}{x-5}+1\Big)\leq 0\\(x-5-1)(x-5+1)>0\end{array}\right$

$\left\{\begin{array}{l}\dfrac{x^2-4x+5}{x-5}\cdot \dfrac{x^2-2x-5}{x-5}\leq 0\\(x-6)(x-4)>0\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{(x^2-4x+5)(x-1+\sqrt6)(x-1-\sqrt6)}{(x-5)^2}\leq 0\\(x-6)(x-4)>0\end{array}\right$

$\left\{\begin{array}{l}x\in [\ 1-\sqrt6\ ;\ 1+\sqrt6\ ]\\x\in (-\infty ;4)\cup (6;+\infty )\end{array}\right\ \ \Rightarrow \ \ x\in [\ 1-\sqrt6\ ;\ 1+\sqrt6\ ]$

$b)\\\left\{\begin{array}{l}log_4\dfrac{(x^2-3x)^2}{(x-5)^2}\geq 0\\log_4(x-5)^2<0\end{array}\right\ \ \left\{\begin{array}{l}\Big(\dfrac{x^2-3x}{x-5}-1\Big)\Big(\dfrac{x^2-3x}{x-5}+1\Big)\geq 0\\(x-5-1)(x-5+1)<0\end{array}\right$

$\left\{\begin{array}{l}\dfrac{(x^2-4x+5)(x-1+\sqrt6)(x-1-\sqrt6)}{(x-5)^2}\geq 0\\(x-6)(x-4)<0\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty ;1-\sqrt6\ ]\cup [\ 1+\sqrt6\, ;\, 5)\cup (\ 5\ ;+\infty \, )\\x\in (\, 4\ ;\ 6\ )\end{array}\right\ \ \Rightarrow \ \ x\in (\ 4;\ 5)\cup (\ 5\ ;\ 6\ )$

$x\in [\ 1-\sqrt6\ ;\ 1+\sqrt6\ ]\cup (\, 4\ ;\ 5\, )\cup (\, 5\ ;\ 6\, )\cup ODZ\ \ \Rightarrow \\\\Otvet:\ x\in [\ 1-\sqrt6\ ;\ 0\ )\cup (\ 3\ ;\ 1+\sqrt6\ ]\cup (\, 4\ ;\ 5\, )\cup (\, 5\ ;\ 6\, )\ .$