Giải bài 9 trang 147 – SGK môn Giải tích lớp 12
Giải các phương trình sau:
a) \({{13}^{2x+1}}-{{13}^{x}}-12=0\);
b) \(\left( {{3}^{x}}+{{2}^{x}} \right)\left( {{3}^{x}}+{{3.2}^{x}} \right)={{8.6}^{x}}\);
c) \({{\log }_{\sqrt{3}}}\left( x-2 \right).{{\log }_{5}}x=2.{{\log }_{3}}\left( x-2 \right)\);
d) \(\log _{2}^{2}x-5{{\log }_{2}}x+6=0\).
a) \({{13}^{2x+1}}-{{13}^{x}}-12=0\);
\(\Leftrightarrow {{13.13}^{2x}}-{{13}^{x}}-12=0\)
Đặt \({{13}^{x}}=t\,\left( t>0 \right)\) phương trình trở thành
\(13{{t}^{2}}-t-12=0 \\ \Leftrightarrow \left[ \begin{aligned} & t=1 \\ & t=-\dfrac{12}{13}\,\left( \text{loại} \right) \\ \end{aligned} \right. \)
Với \(t=1\Leftrightarrow {{13}^{x}}=1\Leftrightarrow x=0\)
Vậy \(S=\left\{ 0 \right\} \)
b) \(\left( {{3}^{x}}+{{2}^{x}} \right)\left( {{3}^{x}}+{{3.2}^{x}} \right)={{8.6}^{x}}\);
\(\Leftrightarrow {{3}^{2x}}+{{4.6}^{x}}+{{3.2}^{2x}}={{8.6}^{x}} \\ \Leftrightarrow {{3}^{2x}}+{{3.2}^{2x}}-{{4.6}^{x}}=0 \)
Chia cả hai vế cho \(6^x\) ta được
\({{\left( \dfrac{3}{2} \right)}^{x}}+3.{{\left( \dfrac{2}{3} \right)}^{x}}-4=0\)
Đặt \({{\left( \dfrac{3}{2} \right)}^{x}}=t,\,\left( t>0 \right)\) phương trình trở thành
\(t+\dfrac{3}{t}-4=0 \\ \Rightarrow {{t}^{2}}-4t+3=0 \\ \Leftrightarrow \left[ \begin{aligned} & t=1 \\ & t=3 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{\left( \dfrac{3}{2} \right)}^{x}}=1 \\ & {{\left( \dfrac{3}{2} \right)}^{x}}=3 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=0 \\ & x={{\log }_{\frac{3}{2}}}3 \\ \end{aligned} \right. \)
Vậy \(S=\left\{ 0;\,{{\log }_{\frac{3}{2}}}3 \right\} \)
c) \({{\log }_{\sqrt{3}}}\left( x-2 \right).{{\log }_{5}}x=2.{{\log }_{3}}\left( x-2 \right)\)
Điều kiện: \(x>2\)
\(\begin{aligned} & \Leftrightarrow 2{{\log }_{3}}\left( x-2 \right).{{\log }_{5}}x-2{{\log }_{3}}\left( x-2 \right)=0 \\ & \Leftrightarrow 2{{\log }_{3}}\left( x-2 \right)\left( {{\log }_{5}}x-1 \right)=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\log }_{2}}\left( x-2 \right)=0 \\ & {{\log }_{5}}x=1 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x-2=1 \\ & x=5 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=3 \\ & x=5 \\ \end{aligned} \right. \end{aligned} \)
Vậy \( S=\left\{ 3;\,5 \right\} \).
d) \(\log _{2}^{2}x-5{{\log }_{2}}x+6=0\)
Điều kiện: \(x>0 \)
Đặt \({{\log }_{2}}x=t\) phương trình trở thành
\({{t}^{2}}-5t+6=0 \\ \Leftrightarrow \left[ \begin{aligned} & t=2 \\ & t=3 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{2}}x=2 \\ & {{\log }_{2}}x=3 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=4 \\ & x=8 \\ \end{aligned} \right. \)
Vậy \(S=\left\{ 4;\,8 \right\}\)