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Δημοσιεύτηκε: **Πέμ Αύγ 07, 2014 8:20 pm**

Να λυθεί το σύστημα:

$\displaystyle{ \left\{ {\begin{array}{*{20}c} {2x^2 - 2x - 10 + \sqrt {x + 2} = 3y^2 + 27y + \sqrt {2y + 5} } \hfill \\ \\ {x^2 - 6x + 11 = 7y - y^2 } \hfill \\\end{array}} \right.}$

Δημοσιεύτηκε: **Πέμ Αύγ 07, 2014 9:01 pm**

Howdy Γιώργος!!
Μετά τους περιορισμούς $x \in \left[ { - 2, + \infty } \right),y \in \left[ { - \frac{5}{2}, + \infty } \right)$ έχουμε:
$\left\{ {\begin{array}{*{20}{c}} {2{x^2} - 2x - 10 + \sqrt {x + 2} = 3{y^2} + 27y + \sqrt {2y + 5} } \\ {} \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right. \Rightarrow$

$\left\{ {\begin{array}{*{20}{c}} {2{x^2} - 2x - 10 + \sqrt {x + 2} = {{\left( {2y + 5} \right)}^2} - {y^2} + 7y - 25 + \sqrt {2y + 5} } \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right. \Rightarrow$

$\left\{ {\begin{array}{*{20}{c}} {2{x^2} - 2x - 10 + \sqrt {x + 2} = {{\left( {2y + 5} \right)}^2} + {x^2} - 6x + 11 - 25 + \sqrt {2y + 5} } \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right. \Rightarrow$

$\left\{ {\begin{array}{*{20}{c}} {{{\left( {x + 2} \right)}^2} + \sqrt {x + 2} = {{\left( {2y + 5} \right)}^2} + \sqrt {2y + 5} } \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right.\mathop \Rightarrow \limits_{b = \sqrt {2y + 5} \geqslant 0}^{a = \sqrt {x + 2} \geqslant 0}$

$\left\{ {\begin{array}{*{20}{c}} {{a^4} + a = {b^4} + b} \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right. \Rightarrow$

$\left\{ {\begin{array}{*{20}{c}} {a = b \vee \left( {a + b} \right)\left( {{a^2} + {b^2}} \right) + 1 = 0\alpha \delta .} \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right. \Rightarrow$

$\left\{ {\begin{array}{*{20}{c}} {\sqrt {x + 2} = \sqrt {2y + 5} } \\ {{x^2} - 6x + 11 = 7y - {y^2}} \end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}} {x = 2y + 3} \\ {5{y^2} - 7y + 2 = 0} \end{array} \Rightarrow } \right.$

$\left\{ {\begin{array}{*{20}{c}} {y = 1} \\ {x = 5} \end{array}} \right. \vee \begin{array}{*{20}{c}} {y = \frac{2}{5}} \\ {x = \frac{{19}}{5}} \end{array}$

Αν έκανα καλά τις πράξεις επαληθεύουν και είναι δεκτές.

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Άρρητο σύστημα (3)

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