Giải bài 63 trang 123 SGK giải tích nâng cao 12
Giải các phương trình sau:
a) \({{\left( 2+\sqrt{3} \right)}^{2x}}=2-\sqrt{3}\); b) \({{2}^{{{x}^{2}}-3x+2}}=4\);
c) \({{2.3}^{x+1}}-{{6.3}^{x-1}}-{{3}^{x}}=9\); d) \({{\log }_{3}}\left( {{3}^{x}}+8 \right)=2+x\).
a) \({{\left( 2+\sqrt{3} \right)}^{2x}}=2-\sqrt{3} \\ \)
Ta có
\(\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=4-3=1 \\ \Leftrightarrow 2-\sqrt{3}=\dfrac{1}{2+\sqrt{3}}={{\left( 2+\sqrt{3} \right)}^{-1}} \)
\(\Rightarrow {{\left( 2+\sqrt{3} \right)}^{2x}}=2-\sqrt{3} \\ \Leftrightarrow 2x=-1 \\ \Leftrightarrow x=-\dfrac{1}{2} \\ b)\, {{2}^{{{x}^{2}}-3x+2}}=4={{2}^{2}} \\ \Leftrightarrow {{x}^{2}}-3x+2=2 \\ \Leftrightarrow {{x}^{2}}-3x=0 \\ \Leftrightarrow \left[ \begin{align} & x=0 \\ & x=3 \\ \end{align} \right. \\ c) \, {{2.3}^{x+1}}-{{6.3}^{x-1}}-{{3}^{x}}=9 \\ \Leftrightarrow {{6.3}^{x}}-{{2.3}^{x}}-{{3}^{x}}=9 \\ \Leftrightarrow {{3.3}^{x}}=9 \\ \Leftrightarrow {{3}^{x}}=3={{3}^{1}} \\ \Leftrightarrow x=1 \)
\(d)\,{{\log }_{3}}\left( {{3}^{x}}+8 \right)=2+x \\ \Leftrightarrow {{3}^{x}}+8={{3}^{2+x}} \\ \Leftrightarrow {{3}^{x}}+8={{9.3}^{x}} \\ \Leftrightarrow {{8.3}^{x}}=8 \\ \Leftrightarrow {{3}^{x}}=1={{3}^{0}} \\ \Leftrightarrow x=0 \)