Question

решите, пожалуйста. спасибо большое​

Answer 1

$\left\{\begin{matrix}2x^2-xy-3y=7\\ 2x^2+x-3=(x-1)(y+5)\end{matrix}\right.\\2x^2+x-3=(x-1)(y+5)\Leftrightarrow (x-1)(2x+3)-(x-1)(y+5)=0\\(x-1)(2x-y-2)=0\\y=2x-2\Rightarrow 2x^2-x(2x-2)-3(2x-2)-7=0\Leftrightarrow 4x=-1\Rightarrow x=-\frac{1}{4}\Rightarrow y=-\frac{5}{2}\\x=1\Rightarrow 5+4y=0\Rightarrow y=-\frac{5}{4}$

Answer 2

Ответ:

$(-\dfrac{1}{4}, -\dfrac{5}{2}), (1, -\dfrac{5}{4})$

Объяснение:

$\left \{ {{2x^2 - xy - 3y = 7} \atop {2x^2 + x - 3 = (x - 1)(y + 5) = xy + 5x - y - 5}} \right.$

$\left \{ {{2x^2 - xy = 3y + 7} \atop {2x^2 - xy = 4x - y - 2}} \right. \Rightarrow 3y + 7 = 4x - y - 2$

$4y = 4x - 9$

$y = x - \dfrac{9}{4}$

$2x^2 - x \cdot (x - \dfrac{9}{4}) = 3 (x - \dfrac{9}{4}) + 7$

$2x^2 - x^2 + \dfrac{9}{4}x = 3x - \dfrac{27}{4} + 7$

$x^2 - \dfrac{3}{4} x - \dfrac{1}{4} = 0$

$x_{1, 2} = \left \{ {{-\frac{1}{4}} \atop {1}} \right.$

$y_{1, 2} = x_{1, 2} - \dfrac{9}{4} = \left \{ {{-\frac{5}{2}} \atop {-\frac{5}{4}}} \right.$