Giải các hệ phương trình:
$1) \left\{ \begin{array}{l}
{{4}^{\frac{{x}}{{y}}{ + }\frac{{y}}{{x}}}}{ = 32                               (1)}\\{lo}{{g}_{3}}{(x + y) = 1 - lo}{{g}_{3}}{(x - y)    
(2)}\end{array} \right.{  }$ 
$2) \left\{ \begin{array}{l}
{{y}^{{2lo}{{g}_{y}}{x}}}{ = 3 + 4y}\\
{lo}{{g}_{x}}{xy = lo}{{g}_{y}}{{x}^{2}}
\end{array} \right.$
 $1$) Ta có:
$(1)  \Leftrightarrow {{2}^{{2}\left( {\frac{{x}}{{y}}{ +
}\frac{{y}}{{x}}} \right)}}{ = }{{2}^{5}}{  }
\Leftrightarrow \frac{{x}}{{y}}{   + }\frac{{y}}{{x}}{  =
}\frac{{5}}{{2}}$
    ${\rm{ }} \Leftrightarrow \frac{{x}}{{y}}{   +
}\frac{{y}}{{x}}{  = }\frac{{5}}{{2}} \Leftrightarrow
\frac{{x}}{{y}}{   = 2 } \vee { }\frac{{x}}{{y}}{ =
}\frac{{1}}{{2}}$
$(1), (2)  \Leftrightarrow \left\{ \begin{array}{l}{x = 2y  \neq    0}\\
{lo}{{g}_{3}}{(x + y) = 1 - lo}{{g}_{3}}{(x - y)}
\end{array} \right.{ } \vee \left\{ \begin{array}{l}
{y = 2x \neq    0}\\
{lo}{{g}_{3}}{(x + y) = 1 - lo}{{g}_{3}}{(x - y)}
\end{array} \right.$
    $ \Leftrightarrow {x = 2,y = 1}$
$2$) Điều kiện: $0<x, y\neq 1$
$\begin{array}{l}
{{2}^{{2lo}{{g}_{y}}{x}}}{ = 3 + 4y} \Leftrightarrow
{{x}^{2}}{ = 3 + 4y}\\
{lo}{{g}_{x}}{xy = lo}{{g}_{y}}{{x}^{2}}
\Leftrightarrow {1 + lo}{{g}_{x}}{xy =
2lo}{{g}_{y}}{x   (*)}
\end{array}$
Đặt: $t = {lo}{{g}_{y}}{x} \Rightarrow
{lo}{{g}_{y}}{x = }\frac{{1}}{{t}}{     (t \neq    0)}$
$\begin{array}{l}
(*) \Leftrightarrow {2}{{t}^{2}}{ - t - 1 = 0} \Leftrightarrow \left[
\begin{array}{l}
{t = 1}\\
{t = - }\frac{{1}}{{2}}
\end{array} \right.\\
{     } \Leftrightarrow {x = y } \vee { x = }\frac{{1}}{{\sqrt {y}
}}
\end{array}$
         $\begin{array}{l}
 \Leftrightarrow \left\{ \begin{array}{l}
{x = y}\\
{{x}^{2}}{ - 4x - 3 = 0}
\end{array} \right.{  } \vee { }\left\{ \begin{array}{l}
{x = }\frac{{1}}{{\sqrt {y} }}\\
\frac{{1}}{{y}}{ = 3 + 4y}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x = 2 + }\sqrt {7} \\
{y = 2 + }\sqrt {7}
\end{array} \right.{  } \vee { }\left\{ \begin{array}{l}
{x = 2}\\
{x = }\frac{{1}}{{4}}
\end{array} \right.
\end{array}$

Thẻ

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