Question

алгебра 9 клас, без графика​

Answer 1

Ответ: (-4/9; -7/3), (1; 2)

Объяснение:

$\left \{ {\big{x+y^2=5,\;\;\quad\quad} \atop \big{2x-y-y^2=-4}} \right.\\\\\\\left \{ {\big{x=5-y^2,\;\quad\quad\quad\quad\quad} \atop \big{2(5-y^2)-y-y^2=-4}} \right.$

Решим отдельно 2-е уравнение:

$2(5-y^2)-y-y^2=-4\\ \\ 10-2y^2-y-y^2+4=0\\\\ 3y^2+y-14=0\\ \\ D=1^2-4\cdot3\cdot(-14)=1+168=169\\ \\ y_1=\dfrac{-1-\sqrt{169} }{2\cdot3} =\dfrac{-1-13}{6}=-\dfrac{7}{3} \\ \\ y_2=\dfrac{-1+\sqrt{169} }{2\cdot3} =\dfrac{-1+13}{6}=2$

Вернёмся к системе:

$\left \{ {\big{x=5-y^2,} \atop {\left \lbrack {\big{y=-\dfrac{7}{3}, } \atop \big{y=2}\quad\;}\;\; \right. }} \right.\\\\\\\left \lbrack {{\left \{ {\big{x=5-\Big(-\dfrac{7}{3} \Big)^2,} \atop \big{y=-\dfrac{7}{3}\quad\;\quad\quad\quad}} \right. } \atop {\left \{ {\big{x=5-2^2,} \atop \big{y=2\;\;\quad\quad}}\quad\quad\;\; \right. }} \right. \\\\\\\left \lbrack {{\left \{ {\big{x=-\dfrac{4}{9},} \atop \big{y=-\dfrac{7}{3}}} \right. } \atop {\left \{ {\big{x=1,} \atop\big{y=2}}\;\;\; \right.$