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$1)$ Điều kiện: $\left\{ \begin{array}{l}
1 \ne x > 0,\,\,y > 0\\
x - 3y > 0
\end{array} \right.$
$(1)\,\,\, \Leftrightarrow \,\,\left\{ \begin{array}{l}
\left( {1 - \frac{2}{5}{{\log }_x}y} \right){\log _x}y = \frac{2}{5}\\
x - 3y = 4
\end{array} \right.$
Đặt $t = {\log _x}y$ ta sẽ có $t = \frac{1}{2}$ hoặc $t = 2$
Với $t = \frac{1}{2}$ suy ra hệ $\left\{ \begin{array}{l}
x = {y^2}\\
x - 3y = 4
\end{array} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x = 16\\
y = 4
\end{array} \right.$
Với $t = 2$ suy ra hệ $\left\{ \begin{array}{l}
y = {x^2}\\
x - 3y = 4
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
y = {x^2}\\
\left[ \begin{array}{l}
x = - 1\,\,\,\,\,\,\,\,\,(l)\\
x = - \frac{4}{3}
\end{array} \right.
\end{array} \right.$
ĐS: $\left( {16,4} \right)$
$2)$
Điều kiện: $\left\{ \begin{array}{l} 1\ne y>0\\ x>0\\ y>3x \end{array} \right.$
Hệ đã cho tương đương với:
$\left\{ \begin{array}{l} \log_x y+\log_y x=\frac 5 2\\ \log_4 (y-3x)=1 \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} \log_x y=2\vee \log_x y=\frac 1 2\\ y-3x=4 \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} y=x^2\\ y=\sqrt{x} \end{array} \right.
\\ y=4+3x \end{array} \right.
$
$\Leftrightarrow \left\{ \begin{array}{l} y=3x+4\\
\left[ \begin{array}{l} x^2-3x-4=0\\ 3x-\sqrt x+4=0 (vô nghiệm)\end{array} \right. \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} y=3x+4\\ x=4 \vee x=-1(loại) \end{array} \right.
\Leftrightarrow \left\{ \begin{array}{l} x=4\\ y=16 \end{array} \right.$
Vậy nghiệm $(x,y)$của hệ là: $\left( {4,16} \right)$
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