Giải bài 2.67 trang 133 - SBT Giải tích lớp 12
Giải các phương trình sau :
\(\begin{align} & a)\,{{9}^{x}}-{{3}^{x}}-6=0. \\ & b)\,\,{{e}^{2x}}-3{{e}^{x}}-4+12{{e}^{-x}}=0; \\ & c)\,\,{{3.4}^{x}}+\dfrac{1}{3}{{.9}^{x+2}}={{6.4}^{x+1}}-\dfrac{1}{2}{{.9}^{x+1}}; \\ & d)\,\,{{2}^{{{x}^{2}}-1}}-{{3}^{{{x}^{2}}}}={{3}^{{{x}^{2}}-1}}-{{2}^{{{x}^{2}}+2}}. \\ \end{align} \)
\(\begin{aligned} & a) \\ & {{9}^{x}}-{{3}^{x}}-6=0 \\ & \Leftrightarrow {{3}^{2x}}-{{3}^{x}}-6=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{3}^{x}}=-2\,\,\,\,(\text{loại}) \\ & {{3}^{x}}=3 \\ \end{aligned} \right. \\ & \Leftrightarrow x=1 \end{aligned} \)
\(\begin{aligned} & b) \\ & {{e}^{2x}}-3{{e}^{x}}-4+12{{e}^{-x}}=0 \\ & \Leftrightarrow {{e}^{2x}}-3{{e}^{x}}-4+\dfrac{12}{{{e}^{x}}}=0 \\ & \Rightarrow {{e}^{3x}}-3{{e}^{2x}}-4{{e}^{x}}+12=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{e}^{x}}=-2\,\,\,\,\,\left( \text{loại} \right) \\ & {{e}^{x}}=2 \\ & {{e}^{x}}=3 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=\ln 2 \\ & x=\ln 3 \\ \end{aligned} \right. \\ \end{aligned} \)
\(\begin{aligned} & c) \\ & {{3.4}^{x}}+\dfrac{1}{3}{{.9}^{x+2}}
={{6.4}^{x+1}}-\dfrac{1}{2}{{.9}^{x+1}} \\ & \Leftrightarrow {{3.4}^{x}}-{{24.4}^{x}}+{{27.9}^{x}}+\dfrac{9}{2}{{.9}^{x}}=0
\\ & \Leftrightarrow \dfrac{63}{2}{{.9}^{x}}={{21.4}^{x}} \\ & \Leftrightarrow {{\left( \dfrac{9}{4} \right)}^{x}}=\dfrac{2}{3} \\ & \Leftrightarrow x=-\dfrac{1}{2} \\ \end{aligned}\)
\(\begin{aligned} & d) \\ & {{2}^{{{x}^{2}}-1}}-{{3}^{{{x}^{2}}}}={{3}^{{{x}^{2}}-1}}-{{2}^{{{x}^{2}}+2}} \\ & \Leftrightarrow \dfrac{1}{2}{{.2}^{{{x}^{2}}}}-{{3}^{{{x}^{2}}}}=\dfrac{1}{3}{{.3}^{{{x}^{2}}}}-{{4.2}^{{{x}^{2}}}} \\ & \Leftrightarrow \dfrac{9}{2}{{.2}^{{{x}^{2}}}}=\dfrac{4}{3}{{.3}^{{{x}^{2}}}} \\ & \Leftrightarrow {{\left( \dfrac{2}{3} \right)}^{{{x}^{2}}}}=\dfrac{8}{27} \\ & \Leftrightarrow {{x}^{2}}=3 \\ & \Leftrightarrow x=\pm \sqrt{3} \\ \end{aligned} \)