Задача 64446 ...

6274e07e93c7c47801bd018b

06.05.2022 11:49:44

Решить неравенство
[m](x-4)(\log_{5}{(125-25x)}-\log_{6}{(x^2+x-6)}+\frac{1}{\log_{(5-x)}{0,2}}+x+6) ≤ x^2+2x-24[/m]

5f3ea7e3faf909182968ddd9

06.05.2022 17:15:15

ОДЗ:
[m]\left\{\begin {matrix}125-5x>0\\x^2+x-6>0\\5-x>0\\5-x ≠1 \end {matrix}\right.[/m] ⇒ [m]\left\{\begin {matrix}x<5\\(x+3)(x-2)>0\\\\x ≠4 \end {matrix}\right.[/m]

[m]x ∈ (- ∞ ;-3)\cup(2;4)\cup(4;5)[/m]

В условиях ОДЗ:

[m]log_{5}(125-5x)=log_{5}25\cdot(5-x)=log_{5}25+log_{5}(5-x)=2+log_{5}(5-x)[/m]

[m]log_{5-x}(0,2)=\frac{1}{log_{0,2}(5-x)}=log_{5^{-1}}(5-x)=-log_{5}(5-x)[/m]

Неравенство принимает вид:

[m](x-4)\cdot (2+log_{5}(5-x)-log_{6}(x^2+x-6)-log_{5}(5-x)+x+6) ≤ x^2+2x-24[/m]

[m](x-4)\cdot (2-log_{6}(x^2+x-6)+x+6) ≤ (x-4)(x+6)[/m]

[m](x-4)\cdot (2-log_{6}(x^2+x-6)+x+6) - (x-4)(x+6) ≤0 [/m]

[m](x-4)\cdot (2-log_{6}(x^2+x-6)+x+6-(x+6)) ≤0 [/m]

[m](x-4)\cdot (2-log_{6}(x^2+x-6)+x+6-x-6) ≤0 [/m]

[m](x-4)\cdot (2-log_{6}(x^2+x-6)) ≤0 [/m]


[m]\left\{\begin {matrix}x-4 ≤ 0\\2-log_{6}(x^2+x-6) ≥ 0 \end {matrix}\right.[/m] или [m]\left\{\begin {matrix}x-4 ≥ 0\\2-log_{6}(x^2+x-6) ≤ 0 \end {matrix}\right.[/m]

[m]\left\{\begin {matrix}x-4 ≤ 0\\log_{6}(x^2+x-6) ≤ 2\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}x-4 ≥ 0\\log_{6}(x^2+x-6) ≥2 \end {matrix}\right.[/m]

[m]\left\{\begin {matrix}x-4 ≤ 0\\x^2+x-6 ≤ 36\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}x-4 ≥ 0\\x^2+x-6 ≥36 \end {matrix}\right.[/m]

[m]\left\{\begin {matrix}x-4 ≤ 0\\(x+7)(x-6) ≤ 0\end {matrix}\right.[/m] или [m]\left\{\begin {matrix}x-4 ≥ 0\\(x+7)(x-6) ≥0
\end {matrix}\right.[/m]


[m]x ∈ [-7;4] [/m] или [m]x ∈ [6;+ ∞ )[/m]

С учетом ОДЗ [m]x ∈ (- ∞ ;-3)\cup(2;4)\cup(4;5)[/m]

получаем ответ:

[m][-7;-3)\cup(2;4)[/m]