A) log1/6(10-x)+log1/6(x-3)>=-1
Б)log1/3(x^2-1) / log0,3 5 <=0
[m]log_{\frac{1}{6}}(10–x)+log_{\frac{1}{6}}(x–3) ≥ –1[/m]
Неравенство имеет смысл при
[m]\left\{\begin {matrix}10-x>0\\x-3>0\end {matrix}\right.[/m]; [m]\left\{\begin {matrix}x<10\\x>3\end {matrix}\right.[/m], т.е при [red] [m] x ∈ (3;10)[/m][/red]
[m]log_{\frac{1}{6}}(10–x)(x–3) ≥ –1[/m]
[m]log_{\frac{1}{6}}(10–x)(x–3) ≥ –1\cdot log_{\frac{1}{6}}\frac{1}{6}[/m]
[m]log_{\frac{1}{6}}(10–x)(x–3) ≥ log_{\frac{1}{6}}(\frac{1}{6})^{-1}[/m]
[m](10–x)(x–3) ≤ (\frac{1}{6})^{-1}[/m]
[m](10–x)(x–3) ≤6[/m] ⇒ [m]x^2-13x+36 ≥ 0[/m]
D=169-144=25
x_(1)=4; x_(2)=9
[m]x^2-13x+36 ≥ 0[/m] ⇒ [m] x ≤ 4[/m] или [m] x ≥ 9[/m]
C учетом [red] [m] x ∈ (3;10)[/m][/red]
получаем о т в е т
[m] x ∈ (3;4]\cup[9;10)[/m]
Б)[m]\frac{log_{\frac{1}{3}}(x^2–1)}{ log_{0,3} 5} ≤ 0[/m]
так как [m] log_{0,3} 5 ≤ log_{0,3} 1=0[/m]
данное неравенство сводится к решению неравенства:
[m]log_{\frac{1}{3}}(x^2–1) ≥ 0[/m] ⇒[m]0<x^2–1 ≤1 [/m] ⇒ [m]\left\{\begin {matrix}x^2-1>0\\x^2-1≤ 1\end {matrix}\right.[/m]; [m]\left\{\begin {matrix}|x|>1\\x^2≤ 2\end {matrix}\right.[/m]; [m]\left\{\begin {matrix}|x|>1\\|x|≤\sqrt{ 2}\end {matrix}\right.[/m]
О т в е т. [-sqrt(2);-1) U(1;sqrt(2)]