Bài 104211

Question

Giải các phương trình:
$a) |x-1|^{\log^2x-\log x^2}=|x-1|^3   (1)$
$b) 5\log x = 60 – x^{\log5}               (2)$
$c) {{3}^{{\log}_{3}^{2}{x}}}{ + }{{x}^{{\log}_{3}{x}}}{ = 162}         (3)$

score 0 · Answer 1

$a)    (1) \Leftrightarrow \left\{ \begin{array}{l}
\left| {{x - 1}} \right|{ > 0}\\
{(}\left| {{x - 1}} \right|{ - 1)(l}{{g}^{2}}{x -
lg}{{x}^{2}}{ - 3) = 0}
\end{array} \right.$
$\begin{array}{l}
\left\{ \begin{array}{l}
\left| {{x - 1}} \right|{ = 1}\\
{x\# 1}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x\# 1}\\
{x > 0}\\
{l}{{g}^{2}}{x - lg}{{x}^{2}}{ - 3 = 0}
\end{array} \right.\\
\Leftrightarrow {x = 2} \vee { }\left\{ \begin{array}{l}
{x \# 1}\\
{lgx = - 1}
\end{array} \right.{ } \vee \left\{ \begin{array}{l}
{x \# 1}\\
{lgx = 3}
\end{array} \right.\\
{x = 2} \vee {x = }\frac{{1}}{{{10}}} \vee {x = 1000}
\end{array}$
$b)$    Nhận xét : ${{3}^{{log}_{3}^{2}{x}}}{ =
}{\left( {{{3}^{{lo}{{g}_{3}}{x}}}}
\right)^{{lo}{{g}_{3}}{x}}}{ =
}{{x}^{{log}_{3}^{x}}}$
$(3) \Leftrightarrow {{x}^{{log}_{3}^{x}}}{ +
}{{x}^{{log}_{3}^{x}}}{ = 162}$
    $\begin{array}{l}
\Leftrightarrow {{x}^{{log}_{3}^{x}}}{ =
}{{3}^{4}} \Leftrightarrow {\left( {{lo}{{g}_{3}}}
\right)^{2}}{ = 4}\\
\Leftrightarrow \left[ \begin{array}{l}
{lo}{{g}_{3}}{x = 2}\\
{lo}{{g}_{3}}{x =- 2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{x = 9}\\
{x = }\frac{{1}}{{9}}
\end{array} \right.
\end{array}$

Bài 104211

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Bài 104211

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