Giải bài 10 trang 147 – SGK môn Giải tích lớp 12
Giải các bất phương trình sau:
a) \(\dfrac{{{2}^{x}}}{{{3}^{x}}-{{2}^{x}}}\le 2\);
b) \({{\left( \dfrac{1}{2} \right)}^{{{\log }_{2}}\left( {{x}^{2}}-1 \right)}}>1\);
c) \({{\log }^{2}}x+3\log x\ge 4\);
d) \(\dfrac{1-{{\log }_{4}}x}{1+{{\log }_{2}}x}\le \dfrac{1}{4}\).
a) \(\dfrac{{{2}^{x}}}{{{3}^{x}}-{{2}^{x}}}\le 2\)
\(\Leftrightarrow \dfrac{{{2}^{x}}}{{{3}^{x}}-{{2}^{x}}}-2\le 0 \\ \Leftrightarrow \dfrac{{{3.2}^{x}}-{{2.3}^{x}}}{{{3}^{x}}-{{2}^{x}}}\le 0 \\ \Leftrightarrow \dfrac{3.{{\left( \dfrac{2}{3} \right)}^{x}}-2}{1-{{\left( \dfrac{2}{3} \right)}^{x}}}\le 0 \)
Đặt \({{\left( \dfrac{2}{3} \right)}^{x}}=t,\,\left( t>0 \right)\) bất phương trình trở thành
\(\dfrac{3t-2}{1-t}\le 0 \\ \Leftrightarrow \left[ \begin{aligned} & t\le \dfrac{2}{3} \\ & t>1 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{\left( \dfrac{2}{3} \right)}^{x}}\le \dfrac{2}{3} \\ & {{\left( \dfrac{2}{3} \right)}^{x}}>1 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x\ge 1 \\ & x<0 \\ \end{aligned} \right. \)
Vậy \(S=\left( -\infty ;\,0 \right)\cup \left[ 1;\,+\infty \right)\)
b) \({{\left( \dfrac{1}{2} \right)}^{{{\log }_{2}}\left( {{x}^{2}}-1 \right)}}>1\)
Điều kiện: \(\left[ \begin{aligned} & x>1 \\ & x<-1 \\ \end{aligned} \right. \)
\(\Leftrightarrow {{\log }_{2}}\left( {{x}^{2}}-1 \right)<0 \\ \Leftrightarrow {{x}^{2}}-1<1 \\ \Leftrightarrow {{x}^{2}}<2 \\ \Leftrightarrow -\sqrt{2}< x < \sqrt{2} \)
Vậy \(S=\left( -\sqrt{2};\,-1 \right)\cup \left( 1;\,\sqrt{2} \right) \)
c) \({{\log }^{2}}x+3\log x\ge 4\)
Điều kiện: \(x>0\)
\(\Leftrightarrow {{\log }^{2}}x+3\log x-4\ge 0\)
Đặt \(\log x=t\) bất phương trình trở thành
\({{t}^{2}}+3t-4\ge 0 \\ \Leftrightarrow \left[ \begin{aligned} & t\ge 1 \\ & t\le -4 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & \log x\ge 1 \\ & \log x\le -4 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x\ge 10 \\ & x\le \dfrac{1}{{{10}^{4}}} \\ \end{aligned} \right. \)
Vậy \(S=\left( 0;\dfrac{1}{{{10}^{4}}} \right]\cup \left[ 10;\,+\infty \right)\)
d) \(\dfrac{1-{{\log }_{4}}x}{1+{{\log }_{2}}x}\le \dfrac{1}{4}\)
Điều kiện: \(\left\{ \begin{aligned} & x>0 \\ & {{\log }_{2}}x\ne -1 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x>0 \\ & x\ne \dfrac{1}{2} \\ \end{aligned} \right. \)
\(\Leftrightarrow \dfrac{1-\dfrac{1}{2}{{\log }_{2}}x}{1+{{\log }_{2}}x}-\dfrac{1}{4}\le 0 \\ \Leftrightarrow \dfrac{3\left( 1-{{\log }_{2}}x \right)}{1+{{\log }_{2}}x}\le 0 \)
Đặt \({{\log }_{2}}x=t\) bất phương trình trở thành
\(\dfrac{3\left( 1-t \right)}{1+t}\le 0 \\ \Leftrightarrow \left[ \begin{aligned} & t<-1 \\ & t\ge 1 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{2}}x<-1 \\ & {{\log }_{2}}x\ge 1 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x<\dfrac{1}{2} \\ & x\ge 2 \\ \end{aligned} \right. \)
Vậy \(S=\left( 0;\,\dfrac{1}{2} \right)\cup \left[ 2;\,+\infty \right)\)