Cristian Calude, proba pe echipe, R.II, P.II
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Laurian Filip
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Cristian Calude, proba pe echipe, R.II, P.II
Sa se determine valoare minima si valoare maxima a expresiei \( E(x,y)=\frac{2x^2-5xy}{x^2+y^2} \), \( x,y \in \mathbb{R}^* \)
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mihai miculita
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\( \mbox{Notam cu: }t=\frac{2x^2-5xy}{x^2+y^2}\Leftrightarrow (2-t).x^2-5xy-t.y^2=0 \Leftrightarrow (2-t).\left(\frac{x}{y}\right)^2-5.\frac{x}{y}-t=0. \ (1) \)
\( \mbox{Ecuatia } \ (1) \ \mbox{ are solutii in } x;y\in \mathbb{R}\Leftrightarrow \triangle=-4t^2+8t+25\ge 0 \Leftrightarrow t\in\left[\frac{2-\sqrt{29}}{2};\frac{2+\sqrt{29}}{2}\right]\Rightarrow \min E(x;y)=\frac{2-\sqrt{29}}{2} \mbox{ si } \max E(x;y)=\frac{2+\sqrt{29}}{2}. \)
\( \mbox{Ecuatia } \ (1) \ \mbox{ are solutii in } x;y\in \mathbb{R}\Leftrightarrow \triangle=-4t^2+8t+25\ge 0 \Leftrightarrow t\in\left[\frac{2-\sqrt{29}}{2};\frac{2+\sqrt{29}}{2}\right]\Rightarrow \min E(x;y)=\frac{2-\sqrt{29}}{2} \mbox{ si } \max E(x;y)=\frac{2+\sqrt{29}}{2}. \)