Алгебра: Решить неравенство( из приложений решений не нужно)

Решить неравенство( из приложений решений не нужно)

Ответ:

ВсеЗнайка717

12.02.2021 11:57

$lg^4(x^2-26)^4-4\, lg^2(x^2-26)^2\leq 240\\\\ODZ:\ \ x^2-260\ \ \to \ \ x^2-26\ne 0\ \ ,\ \ x\ne \pm \sqrt{26}\ \ \ (\sqrt{26}\approx 5,099)\\\\\star \ \ \log_{a}x^2=2\, \log_{a}|x|\ \ ,\ \ \log_{a}x^4=4\, \log_a}|x|\ \ \star \\\\\Big(lg(x^2-26)^4\Big)^4-4\, \Big(lg(x^2-26)^2\Big)^2\leq 240\\\\\Big(lg(x^2-26)^4\Big)^4-4\, \Big(lg(x^2-26)^2\Big)^2\leq 240\\\\\Big(4\, lg|x^2-26|\Big)^4-4\, \Big(2\, lg|x^2-26|\Big)^2\leq 240\\\\256\cdot lg^4|x^2-26|-16\cdot lg^2|x^2-26|-240\leq 0\ \Big|:16$

$16\cdot lg^4|x^2-26|-lg^2|x^2-26|-15\leq 0\\\\t=lg^2|x^2-26|\geq 0\ \ \ \to \ \ \ 16t^2-t-15\leq 0\ \ ,\ \ D=961=31^2\ ,\\\\t_1=-\dfrac{15}{16}\ \ ,\ \ t_2=1\ \ ,\ \ \ \ (t-1)(t+\dfrac{15}{16})\leq 0\ \ ,\\\\znaki:\ \ \ +++[-\dfrac{15}{16}\, ]---[\ 1\ ]+++\ \ ,\ \ \ t\in [-\dfrac{15}{16}\ ;\ 1\ ]\ \ \ ,\ \ t\geq 0\ \ \Rightarrow$

$0\leq t\leq 1\ \ \Rightarrow \ \ \ 0\leq lg^2|x^2-26|\leq 1\ \ \ \Rightarrow \ \ \ \ \left\{\begin{array}{l}lg^2|x^2-26|\geq 0\\lg^2|x^2-26|\leq 1\end{array}\right\\\\\\1)\ \ lg^2|x^2-26|\geq 0\ \ \ \Rightarrow \ \ \ x\ne \pm \sqrt{26}\ \ .\\\\2)\ \ lg^2|x^2-26|\leq 1\ \ \ \ \Rightarrow \ \ \ \lg^2|x^2-26|-1\leq 0\ \ ,\\\\\Big(lg|x^2-26|-1\Big)\Big(lg|x^2-26|+1\Big)\leq 0\ \ \Rightarrow \ \ -1\leq lg|x^2-26|\leq 1$

$\left\{\begin{array}{l}lg|x^2-26|\geq -1\\lg|x^2-26|\leq 1\end{array}\right\ \ \left\{\begin{array}{l}|x^2-26|\geq 10^{-1\\|x^2-26|\leq 10}\end{array}\right\ \ \left\{\begin{array}{l}|x^2-26|\geq 0,1\\-10\leq x^2-26\leq 10\end{array}\right$

$\left\{\begin{array}{l}\left[\begin{array}{l}x^2-26\geq 0,1\\x^2-26\leq -0,1\end{array}\right\\16\leq x^2\leq 36\end{array}\right\ \ \ \left\{\begin{array}{l}\left[\begin{array}{l}x^2\geq 26,1\\x^2\leq 25,9\end{array}\right\\\left\{\begin{array}{l}x^2\geq 16\\x^2\leq 36\end{array}\right\end{array}\right\ \ \ \left\{\begin{array}{l}\left[\begin{array}{l}x^2-26,1\geq 0\\x^2-25,9\leq 0\end{array}\right\\\left\{\begin{array}{l}x^2-16\geq 0\\x^2-36\leq 0\end{array}\right\end{array}\right$

$\left\{\begin{array}{l}\left[\begin{array}{l}x\in (-\infty ;-\sqrt{26.1}\, ]\cup [\, \sqrt{26,1}\, ;+\infty )\\x\in [-\sqrt{25,9}\, ;\, \sqrt{25,9}\, ]\end{array}\right\\\\\left\{\begin{array}{l}x\in (-\infty;-4\, ]\cup [\, 4\, ;+\infty )\\x\in [\, -6\, ;\, 6\, ]\end{array}\right\end{array}\right\\\\\\\left\{\begin{array}{l}x\in (-\infty \, ;-\sqrt{26,1}\, ]\cup [-\sqrt{25,9}\, ;\, \sqrt{25,9}\, ]\cup [\, \sqrt{26,1}\, ;+\infty )\\\\x\in [-6\ ;-4\ ]\cup [\ 4\ ;\ 6\ ]\end{array}\right$