Giải các phương trình sau:
$a) \frac{{1}}{{2}}{log(5x - 4) + log}\sqrt {{x + 1}} { = 2 + log0,18   (1)}$
$b) \sqrt {{lo}{{g}_{{0}{.04}}}{x + 1}} + \sqrt{log_{0.2}x+3}=1  (2)$
$c) log(logx) + log(logx^3 – 2) = 0            (3)$
$a$) Ðiều kện: $x > \frac{4}{5}$
$(1)  \Leftrightarrow {lg(5x - 4) + lg(x + 1) = 4 + 2log0,18}$
     $\begin{array}{l}
 \Leftrightarrow {lg(5}{{x}^{2}}{ + x - 4) = log}{\left( {{0,18}}
\right)^{2}}\\
 \Leftrightarrow {x = 8}
\end{array}$
$b$)    Ðiều kiện $x >0$
Nhận xét: $(0,2)^2 = 0,04,  log_{0,04} x = \frac{1}{2}log_{0,2} x$
Ðặt: $t = log_{0,2} x$
$(2)  \Leftrightarrow \sqrt {{1 + }\frac{{1}}{{2}}{t}} { + }\sqrt
{{3 + t}} { = 1}$
        $\begin{array}{l}
 \Leftrightarrow \sqrt {2} {.}\sqrt {{t + 3}} { = }\sqrt {2} {
- }\sqrt {{2 + t}} \\
 \Leftrightarrow \left\{ \begin{array}{l}
{t} \le {0}\\
{t + 2 = 0}
\end{array} \right. \Leftrightarrow {t = - 2}
\end{array}$
       $ \Leftrightarrow {lo}{{g}_{{0,2}}}{x = - 2   }
\Leftrightarrow {x = 25}$
$c)$ Ðiều  kiện  $lgx > 0$
$(3)  \Leftrightarrow  lg[ lgx(lgx^3 – 2)] = 0    \Leftrightarrow   lgx(3lgx^3 – 2) = 1$
       $ \Leftrightarrow  3( lgx)^2 – 2lgx – 1  = 0    \Leftrightarrow   lgx = 1   \Leftrightarrow   x = 10$

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