Giải bài 2.71 trang 134 - SBT Giải tích lớp 12

a)

ĐK: \(x> 0\)

\(\begin{aligned} & \dfrac{\ln x+2}{\ln x-1}<0 \\ & \Leftrightarrow -2< \ln x<1 \\ & \Leftrightarrow {{e}^{-2}}< x< e \\ \end{aligned} \)

b)

ĐK: \(x> 0\)

\(\begin{aligned} & \log _{0,2}^{2}x-{{\log }_{0,2}}x-6\le 0 \\ & \Leftrightarrow -2\le {{\log }_{0,2}}x\le 3 \\ & \Leftrightarrow 0,{{2}^{3}}\le x\le 0,{{2}^{-2}} \\ \end{aligned} \)

c)

ĐK: \( \left\{ \begin{aligned} & {{x}^{2}}-x-2>0 \\ & 3-x>0 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \left[ \begin{aligned} & x<-1 \\ & x>2 \\ \end{aligned} \right. \\ & x<3 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x<-1 \\ & 2< x<3 \\ \end{aligned} \right. \)

\(\begin{aligned} & \log \left( {{x}^{2}}-x-2 \right)<2\log \left( 3-x \right) \\ & \Leftrightarrow \log \left( {{x}^{2}}-x-2 \right)<\log {{\left( 3-x \right)}^{2}} \\ & \Leftrightarrow {{x}^{2}}-x-2<9-6x+{{x}^{2}} \\ & \Leftrightarrow 5x-11<0 \\ & \Leftrightarrow x<\dfrac{11}{5} \\ \end{aligned} \)

Kết hợp điều kiện: \(x\in \left( -\infty ;-1 \right)\cup \left( 2;\dfrac{11}{5} \right) \)

d)

\(\begin{aligned} & \ln \left| x-2 \right|+\ln \left| x+4 \right|\le 3\ln 2 \\ & \Leftrightarrow \ln \left| \left( x-2 \right)\left( x+4 \right) \right|\le \ln 8 \\ & \Leftrightarrow \left| {{x}^{2}}+2x-8 \right|\le 8 \\ & \Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}+2x-8\ge -8 \\ & {{x}^{2}}+2x-8\le 8 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & {{x}^{2}}+2x\ge 0 \\ & {{x}^{2}}+2x-16\le 0 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x\le -2 \,\text{ hoặc}\, x\ge 0 \\ & -1-\sqrt{17}\le x\le -1+\sqrt{17} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & -1-\sqrt{17} \le x \le -2 \\ &0 \le x \le -1+\sqrt{17} \\ \end{aligned} \right. \\ \end{aligned} \)