Giải bài 68 trang 124 SGK giải tích nâng cao 12

$\begin{aligned} a)\,& {{3}^{x+1}}+{{18.3}^{-x}}=29 \\ & \Leftrightarrow {{3.3}^{x}}+\dfrac{18}{{{3}^{x}}}=29 \\ & \Leftrightarrow {{3.3}^{2x}}-{{29.3}^{x}}+18=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{3}^{x}}=9 \\ & {{3}^{x}}=\dfrac{2}{3} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=2 \\ & x={{\log }_{3}}\dfrac{2}{3}={{\log }_{3}}2-1 \\ \end{aligned} \right. \\ \end{aligned}$

Vậy $S=\left\{ 2;{{\log }_{3}}2-1 \right\}$

$b)\,{{27}^{x}}+{{12}^{x}}={{2.8}^{x}} \\ \Leftrightarrow {{3}^{3x}}+{{2}^{2x}}{{.3}^{x}}={{2.2}^{3x}}$

Chia cả hai vế cho ${{2}^{3x}}$ ta được:

${{\left( \dfrac{3}{2} \right)}^{3x}}+{{\left( \dfrac{3}{2} \right)}^{x}}-2=0$

Đặt $t={{\left( \dfrac{3}{2} \right)}^{x}}$ phương trình trở thành:

$\Leftrightarrow {{t}^{3}}+t-2=0 \\ \Leftrightarrow \left( t-1 \right)\left( {{t}^{2}}+t+2 \right)=0 \\ \Leftrightarrow \left[ \begin{aligned} & t=1 \\ & {{t}^{2}}+t+2=0\,\left( \text{vô nghiệm} \right) \\ \end{aligned} \right. \\ $

Với $t=1\Rightarrow {{\left( \dfrac{3}{2} \right)}^{x}}=1\Leftrightarrow x=0$

Vậy $S=\left\{ 0 \right\}$