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$1)\,\,(1) \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
xy > 0\\
y \ne 0\\
xy = {2^5}\\
\frac{x}{y} = \frac{1}{2}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}
y = 2x\\
2{x^2} = {2^5}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 4\\
y = 8
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 4\\
y = - 8
\end{array} \right.
\end{array} \right.$
Vậy hệ phương trình có 2 nghiệm là ${(4;8);(-4;-8)}$
$2)\,\,(2)\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x > 0,\,\,\,x \ne 1\\
y > 0\\
xy = 64\\
y = {x^5}
\end{array} \right.$
$\Leftrightarrow$ $\left\{ \begin{array}{l}
x^6 = 64\\
y = x^5\\
x>0
\end{array} \right.$ $\Leftrightarrow$ $\left\{ \begin{array}{l}
x = 2\\
y = 32
\end{array} \right.$
Hệ phương trình có nghiệm duy nhất $(2;32)$
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