Слишком простой пример, можно было и самостоятельно посчитать
Х-у=3 (1)
2у²+х²=9 (2)
у=х-3 (1)
2(х-3)²+х²=9 (2)
у=х-3 (1)
2(х²-6х+9)+х²-9=0 (2)
у=х-3 (1)
2х²-12х+18+х²-9=0 (2)
у=х-3 (1)
3х²-12х+9=0
х²-4х+3=0
D=4
x=1
x=3
y=-2
y=-5
Ответ: (1;-2)(3;-5)
Область определения: -1-x≠0 ⇒ x≠-1
Преобразуем:
![\sf y=\dfrac{(x^2+4)(x+1)}{-1-x}=\dfrac{(x^2+4)(x+1)}{-(x+1)}=-x^2-4](https://tex.z-dn.net/?f=%5Csf+y%3D%5Cdfrac%7B%28x%5E2%2B4%29%28x%2B1%29%7D%7B-1-x%7D%3D%5Cdfrac%7B%28x%5E2%2B4%29%28x%2B1%29%7D%7B-%28x%2B1%29%7D%3D-x%5E2-4)
Графиком функции является парабола с ветвями вниз.
Точки для построения: (-3; -13), (-2; -8), (-1; -5) - выколота, (0; -4), (1; -5), (2; 8), (3; -13).
![s=-1/6t^3+1/2t^2+1/2t+1\\ s'=-3/6t^2+2/2t+1/2=0\\ -1/2t^2+t+1/2=0\\ t^2-2t-1=0\\ D=4+4=8\\ t _{12} = \frac{2+- \sqrt{8} }{2} =1+-\sqrt{2}\\ s(1+\sqrt {2})=-1/6(1+\sqrt{2})^3+1/2(1+\sqrt{2})^2+1/2(1+\sqrt{2})+1=\\ -1/6(1+3\sqrt{2}+6+2\sqrt{2})+1/2(1+2+2\sqrt{2})+1/2(1+\sqrt{2})+1=\\ -7/6-5\sqrt{2} /6+3/2+\sqrt{2}+1/2+\sqrt{2}/2+1=3-7/6+4\sqrt{2}/6](https://tex.z-dn.net/?f=s%3D-1%2F6t%5E3%2B1%2F2t%5E2%2B1%2F2t%2B1%5C%5C+s%27%3D-3%2F6t%5E2%2B2%2F2t%2B1%2F2%3D0%5C%5C+-1%2F2t%5E2%2Bt%2B1%2F2%3D0%5C%5C+t%5E2-2t-1%3D0%5C%5C+D%3D4%2B4%3D8%5C%5C+t+_%7B12%7D+%3D+%5Cfrac%7B2%2B-+%5Csqrt%7B8%7D+%7D%7B2%7D+%3D1%2B-%5Csqrt%7B2%7D%5C%5C+s%281%2B%5Csqrt+%7B2%7D%29%3D-1%2F6%281%2B%5Csqrt%7B2%7D%29%5E3%2B1%2F2%281%2B%5Csqrt%7B2%7D%29%5E2%2B1%2F2%281%2B%5Csqrt%7B2%7D%29%2B1%3D%5C%5C+-1%2F6%281%2B3%5Csqrt%7B2%7D%2B6%2B2%5Csqrt%7B2%7D%29%2B1%2F2%281%2B2%2B2%5Csqrt%7B2%7D%29%2B1%2F2%281%2B%5Csqrt%7B2%7D%29%2B1%3D%5C%5C+-7%2F6-5%5Csqrt%7B2%7D+%2F6%2B3%2F2%2B%5Csqrt%7B2%7D%2B1%2F2%2B%5Csqrt%7B2%7D%2F2%2B1%3D3-7%2F6%2B4%5Csqrt%7B2%7D%2F6)
может где в вычислениях ошиблась, и там до конца нужно досчитать