Đáp án:
\(x=3\) và \(x=9.\)
Giải thích các bước giải:
\(\begin{array}{l}
\log _3^2x + \left( {x - 12} \right){\log _3}x + 11 - x = 0\,\,\,\left( * \right)\\
DK:\,\,\,x > 0\\
\left( * \right) \Leftrightarrow \log _3^2x - {\log _3}x + \left( {x - 11} \right){\log _3}x - \left( {x - 11} \right) = 0\\
\Leftrightarrow {\log _3}x\left( {{{\log }_3}x - 1} \right) + \left( {x - 11} \right)\left( {{{\log }_3}x - 1} \right) = 0\\
\Leftrightarrow \left( {{{\log }_3}x + x - 11} \right)\left( {{{\log }_3}x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\log _3}x = 1\\
{\log _3}x = 11 - x
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\,\,\,\,\left( {tm} \right)\\
{\log _3}x = 11 - x\,\,\,\,\left( 1 \right)
\end{array} \right.\\
Xet\,\,\,\left( 1 \right):\,\,\,{\log _3}x = 11 - x\\
Dat\,\,f\left( x \right) = {\log _3}x;\,\,\,\,g\left( x \right) = 11 - x\\
\Rightarrow f'\left( x \right) = \frac{1}{{x\ln 3}} > 0\,\,\forall x > 0\\
\Rightarrow f\left( x \right)\,\,DB\,\,\,tren\,\,\left( {0; + \infty } \right).\\
g'\left( x \right) = - 1 < 0\\
\Rightarrow g\left( x \right)\,\,\,\,Nb\,\,\,tren\,\,\left( {0; + \infty } \right)\\
\Rightarrow pt\,\,\,\left( 1 \right)\,\,\,co\,\,\,nghiem\,\,\,thi\,\,\,co\,\,\,nghiem\,\,\,duy\,\,\,nhat.\\
Ma\,\,f\left( 9 \right) = {\log _3}9 = 2;\,\,\,g\left( 9 \right) = 11 - 9 = 2\\
\Rightarrow x = 9\,\,\,la\,\,\,nghiem\,\,\,cua\,\,pt.\\
Vay\,\,\,pt\,\,\,co\,\,hai\,\,\,nghiem:\,\,\,x = 3;\,\,\,x = 9.
\end{array}\)
