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$1)\,\,\,\,\left\{ \begin{array}{l}
x + y > 0\\
x \ne y
\end{array} \right.$ đặt $\begin{array}{l}
u = x + y\\
v = x - y
\end{array}$
Ta có $\left\{ \begin{array}{l}
{u^{\frac{1}{v}}} = \frac{1}{{4\sqrt 3 }}\\
u{.2^{ - v}} = 48
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ \begin{array}{l}
\frac{1}{v}{\log _2}u = - {\log _2}4\sqrt 3 \\
{\log _2}u - v = {\log _2}48
\end{array} \right.$
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
{\log _2}u = v + 4 + {\log _2}3\\
\frac{1}{v}\left( {v + 4 + {{\log }_2}3} \right) = - 2 - \frac{1}{2}{\log _2}3
\end{array} \right.\\
\Leftrightarrow \,\left\{ \begin{array}{l}
v = - 2\\
{\log _2}u = 2 + {\log _2}3
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
v = - 2\\
u = 12
\end{array} \right.
\end{array}$
Do đó ta có hệ $\left\{ \begin{array}{l}
x + y = 12\\
x - y = - 2
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \left\{ \begin{array}{l}
x = 5\\
y = 7
\end{array} \right.$
Hệ đã cho có nghiệm $(x;y)=(5;7)$
$2)$ Biến đổi hệ thành:
$\left\{ \begin{array}{l}
{x^2} = {9^{\frac{2}{y}}}\\
{\left( {\frac{{324}}{{81}}} \right)^{\frac{1}{y}}} = 2
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,(y \ne 0)\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = {9^{\frac{2}{y}}}\\
{2^{\frac{2}{y}}} = 2
\end{array} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x = \pm 3\\
y = 2
\end{array} \right.$
Hệ có $2$ nghiệm $( - 3,2);\,(3,2)$
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Đăng bài 26-04-12 03:58 PM
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