от ammornil » 03 Яну 2013, 20:59
[tex]\left| x+y=5 \\ \frac{1}{x}+\frac{1}{y}=\frac{5}{6} \right, \\
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\cyr{DM}: \left|x \ne 0 \\ y \ne 0 \right, \\
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\left|x+y=5 \\ 6.y+6.x=5xy \right, \hspace{12} \Rightarrow \left|x+y=5 \\ 6.\underbrace{(x+y)}_{5}=5xy \right, \hspace{12} \Rightarrow \left|x=5-y \\ xy=6 \right, \hspace{12} \Rightarrow \left|x=5-y \\ (5-y).y=6 \right, \hspace{12} \Rightarrow \\
\left|x=5-y \\ 5y-y^2-6=0 \right, \hspace{12} \Rightarrow \left|x=5-y \\ y^2-5y+6=0 \right, \hspace{12} \Rightarrow \left|x=5-y \\ (y-2).(y-3)=0 \right, \hspace{12} \Rightarrow \\
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\left|x=5-y \\ y=2 \right, \hspace{12} \cup \hspace{12} \left|x=5-y \\ y=3 \right, \\
\left|x=3 \\ y=2 \right, \hspace{40} \cup \hspace{12} \left|x=2 \\ y=3 \right, \\[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]