Question

помогите пожалуйста ​

Answer 1

Ответ:

$\boxed{(2;\sqrt{21}), (2;-\sqrt{21}),(-3;4), (-3;-4)}$

Объяснение:

$\displaystyle \left \{ {{ 5x^{2} - y^{2} + 6x = 11 } \atop {x^{2}+y^{2} =25}} \right$   $\displaystyle \left \{ {{ 5x^{2} + 6x - 11 = y^{2} } \atop {y^{2} =25 - x^{2} }} \right \Longrightarrow 5x^{2} + 6x - 11 = 25 - x^{2}$

$5x^{2} + 6x - 11 = 25 - x^{2}$

$6x^{2} + 6x - 36 = 0|:6$

$x^{2} + x - 6= 0$

$D = 1 - 4 \cdot 1 \cdot (-6) = 1 + 24 = 25 = 5^{2}$

$x_{1} = \dfrac{-1 + 5}{2} = \dfrac{4}{2} = 2$

$x_{2} = \dfrac{-1 - 5}{2} = \dfrac{-6}{2} = -3$

$y^{2} =25 - x^{2}$

$\sqrt{y^{2}} = \sqrt{25 - x^{2}}$

$|y| = \sqrt{25 - x^{2} }$

$y = б\sqrt{25 - x^{2} }$

$x_{1} = 2$

$y_{1} = \sqrt{25 - x_{1}^{2} } = \sqrt{25 - 2^{2} } = \sqrt{25 -4} = \sqrt{21}$

$y_{2} = -\sqrt{25 - x_{1}^{2} } = -\sqrt{25 - 2^{2} } = -\sqrt{25 -4} = -\sqrt{21}$

$(2;\sqrt{21}), (2;-\sqrt{21})$

$x_{2} = -3$

$y_{3} = \sqrt{25 - x_{2}^{2} } = \sqrt{25 - (-3)^{2} } = \sqrt{25 -9} = \sqrt{16} = 4$

$y_{4} = -\sqrt{25 - x_{1}^{2} } = -\sqrt{25 - (-3)^{2} } = -\sqrt{25 - 9} = -\sqrt{16} = -4$

$(-3;4), (-3;-4)$