$\begin{cases} x + y = 3 \\ xz + yt = 5 \\ xy^{2} + yt^{2} = 41 \\ xz^{3} + yt^{3} = 121 \end{cases}$
Giải hệ 4 phương trình 4 ẩn.
Started By Valar Morghulis, 29-06-2013 - 10:43
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Posted 29-06-2013 - 10:43
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