Giải bài 97 trang 132 SGK giải tích nâng cao 12
Giải các bất phương trình:
a) \(\dfrac{1-{{\log }_{4}}x}{1+{{\log }_{2}}x}\le \dfrac{1}{2};\) b) \({{\log }_{\frac{1}{\sqrt{5}}}}\left( {{6}^{x+1}}-{{36}^{x}} \right)\ge -2;\)
c) \({{\log }_{\frac{1}{5}}}\left( {{x}^{2}}-6x+18 \right)+2{{\log }_{5}}\left( x-4 \right)<0 \).
a) Điều kiện: \(x>0;x\ne \dfrac{1}{2}\)
\(\dfrac{1-{{\log }_{4}}x}{1+{{\log }_{2}}x}\le \dfrac{1}{2}\Leftrightarrow \dfrac{1-\dfrac{1}{2}{{\log }_{2}}x}{1+{{\log }_{2}}x}-\dfrac{1}{2}\le 0 \\ \Leftrightarrow \dfrac{1-2{{\log }_{2}}x}{2\left( 1+{{\log }_{2}}x \right)}\le 0 \\ \Leftrightarrow \left\{ \begin{aligned} & {{\log }_{2}}x\ge \dfrac{1}{2} \\ & {{\log }_{2}}x<-1 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x\ge \sqrt{2} \\ & x<\dfrac{1}{2} \\ \end{aligned} \right. \)
Vậy \(S=\left( 0;\dfrac{1}{2}\right)\cup \left[\sqrt{2};+\infty \right)\).
b) Điều kiện: \({{6}^{x+1}}-{{36}^{x}}>0\Leftrightarrow {{6}^{x}}<6\Leftrightarrow x<1\)
\({{\log }_{\frac{1}{\sqrt{5}}}}\left( {{6}^{x+1}}-{{36}^{x}} \right)\ge -2 \\ \Leftrightarrow {{6}^{x+1}}-{{36}^{x}}\le 5 \\ \Leftrightarrow {{6}^{2x}}-{{6.6}^{x}}+5\ge 0 \\ \Leftrightarrow \left[ \begin{aligned} & {{6}^{x}}\le1 \\ & {{6}^{x}}\ge5 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x\le0 \\ & x\ge{{\log }_{6}}5 \\ \end{aligned} \right. \)
Vậy \(S=\left(-\infty; 0\right]\cup \left[{{\log }_{6}}5; 1 \right)\)
c) Điều kiện: \(x>4\)
\({{\log }_{\frac{1}{5}}}\left( {{x}^{2}}-6x+18 \right)+2{{\log }_{5}}\left( x-4 \right)<0 \\ \Leftrightarrow {{\log }_{5}}{{\left( x-4 \right)}^{2}}<{{\log }_{5}}\left( {{x}^{2}}-6x+18 \right) \\ \Leftrightarrow {{x}^{2}}-8x+16<{{x}^{2}}-6x+18 \\ \Leftrightarrow -2x<2 \\ \Leftrightarrow x>-1 \)
Vậy \(S=\left( 4;+\infty \right)\)