Giải bài 8 trang 90 – SGK môn Giải tích lớp 12
Giải các bất phương trình:
a) \({{2}^{2x-1}}+{{2}^{2x-2}}+{{2}^{2x-3}}\ge 448;\)
b) \({{\left( 0,4 \right)}^{x}}-{{\left( 2,5 \right)}^{x+1}}>1,5;\)
c) \({{\log }_{3}}\left[ {{\log }_{\frac{1}{2}}}\left( {{x}^{2}}-1 \right) \right]<1;\)
d) \(\log _{0,2}^{2}x-5{{\log }_{0,2}}x<-6.\)
a) \({{2}^{2x-1}}+{{2}^{2x-2}}+{{2}^{2x-3}}\ge 448;\)
\(\begin{align} & \Leftrightarrow \dfrac{1}{2}{{.2}^{2x}}+\dfrac{1}{4}{{.2}^{2x}}+\dfrac{1}{8}{{.2}^{2x}}\ge 448 \\ & \Leftrightarrow \dfrac{7}{8}{{.2}^{2x}}\ge 448 \\ & \Leftrightarrow {{2}^{2x}}\ge 512={{2}^{9}} \\ & \Leftrightarrow 2x\ge 9\Leftrightarrow x\ge \dfrac{9}{2}\\ \end{align} \)
Vậy \(S=\left[ \dfrac{9}{2};\,+\infty \right)\)
b) \({{\left( 0,4 \right)}^{x}}-{{\left( 2,5 \right)}^{x+1}}>1,5;\)
\(\begin{align} & \Leftrightarrow {{\left( \dfrac{2}{5} \right)}^{x}}-{{\left( \dfrac{5}{2} \right)}^{x+1}}>\dfrac{3}{2} \\ & \Leftrightarrow {{\left( \dfrac{2}{5} \right)}^{x}}-\dfrac{5}{2}\dfrac{1}{{{\left( \dfrac{2}{5} \right)}^{x}}}-\dfrac{3}{2}>0 \\ & \Leftrightarrow {{\left( \dfrac{2}{5} \right)}^{2x}}-\dfrac{3}{2}{{\left( \dfrac{2}{5} \right)}^{x}}-\dfrac{5}{2}>0 \\ \\ \end{align} \)
Đặt \({{\left( \dfrac{2}{5} \right)}^{x}}=t>0\) bất phương trình trở thành:
\(\begin{align} & \Leftrightarrow {{t}^{2}}-\dfrac{3}{2}t-\dfrac{5}{2}>0 \\ & \Leftrightarrow \left[ \begin{aligned} & t>\dfrac{5}{2} \\ & t<-1\,\left(\text{loại do}\,t>0 \right) \\ \end{aligned} \right. \\\end{align} \)
Với \(t>\dfrac{5}{2}\Leftrightarrow {{\left( \dfrac{2}{5} \right)}^{x}}>\dfrac{5}{2}={{\left( \dfrac{2}{5} \right)}^{-1}}\Leftrightarrow x<-1\)
Vậy \(S=\left( -\infty ;\,-1 \right)\)
c) \({{\log }_{3}}\left[ {{\log }_{\frac{1}{2}}}\left( {{x}^{2}}-1 \right) \right]<1;\) (1)
Điều kiện: \(\,\left\{ \begin{aligned} & {{\log }_{\frac{1}{2}}}\left( {{x}^{2}}-1 \right)>0 \\ & {{x}^{2}}-1>0 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}-1<1 \\ & {{x}^{2}}-1>0 \\ \end{aligned} \right.\\\Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}<2 \\ & {{x}^{2}}>1 \\ \end{aligned} \right.\Leftrightarrow x\in \left( -\sqrt{2};\,-1 \right)\cup \left( 1;\,\sqrt{2} \right) \)
\(\begin{aligned} & (1) \Leftrightarrow {{\log }_{\frac{1}{2}}}\left( {{x}^{2}}-1 \right)<3 \\ & \Leftrightarrow {{x}^{2}}-1>\dfrac{1}{8} \\ & \Leftrightarrow {{x}^{2}}>\dfrac{9}{8} \\ & \Leftrightarrow x\in \left( -\infty ;\,-\dfrac{3}{2\sqrt{2}} \right)\cup \left( \dfrac{3}{2\sqrt{2}};\,+\infty \right)\\\end{aligned} \)
Vậy \(S=\left( -\sqrt{2};\,-\dfrac{3}{2\sqrt{2}} \right)\cup \left( \dfrac{3}{2\sqrt{2}};\,\sqrt{2} \right)\)
d) \(\log _{0,2}^{2}x-5{{\log }_{0,2}}x<-6.\)
Điều kiện: \(x>0 \)
Đặt \({{\log }_{0,2}}x=t\) bất phương trình trở thành:
\(\begin{aligned} & {{t}^{2}}-5t+6<0\Leftrightarrow \left\{ \begin{aligned} & t>2 \\ & t<3 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{\log }_{0,2}}x>2 \\ & {{\log }_{0,2}}x<3 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & x<0,{{2}^{2}} \\ & x>0,{{2}^{3}} \\ \end{aligned} \right.\\&\Leftrightarrow \left\{ \begin{aligned} & x<0,04 \\ & x>0,008 \\ \end{aligned} \right.\\\end{aligned} \)
Vậy \(S=\left( 0,008;\,0,04 \right) \)