Giải hệ: $ \left\{ \begin{array}{l}
3x^2 + 5xy - 4y^2 = 38   (1)\\
5x^2 - 9xy - 3y^2 = 15   (2)
\end{array} \right. $
Khử  $ {x^2} $  giữa (1) và(2)  $  \Rightarrow 52xy - 11{y^2} = 145 $
$  \Leftrightarrow x = \frac{{11{y^2} + 145}}{{52y}} $    (vì  $ y \ne 0 $ )
Thay vào (1)  $  \Rightarrow 3{\left( {\frac{{11{y^2} + 145}}{{52y}}} \right)^2} + \frac{{52{y^2} + 725}}{{52y}}y - 4{y^2} = 38 $  $  \Leftrightarrow 2531{y^4} + 18494{y^2} - 21025 = 0 $
Đặt  $ {y^2} = t \ge 0 $  ta có: $ 2531{t^2} + 18494t - 21025 = 0 $
Phương trình có nghiệm  $ {t_1} = 1;{t_2} < 0 $
Do đó ta có: $ y =  \pm 1 \Rightarrow x =  \pm 3 $
Vậy: hệ có 2 nghiệm là:  $ \left( {3,1} \right);\left( { - 3, - 1} \right) $

Thẻ

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