Вроде,так..............................
![( x^{2} +3x-1) ^{2} -12 x^{2} -36x+39=0\\\\( x^{2} +3x-1) ^{2} -12( x^{2} +3x)+39=0](https://tex.z-dn.net/?f=%28%20x%5E%7B2%7D%20%2B3x-1%29%20%5E%7B2%7D%20-12%20x%5E%7B2%7D%20-36x%2B39%3D0%5C%5C%5C%5C%28%20x%5E%7B2%7D%20%2B3x-1%29%20%5E%7B2%7D%20-12%28%20x%5E%7B2%7D%20%2B3x%29%2B39%3D0)
Обозначим x² + 3x - 1 = m , тогда x² + 3x = m + 1
m² - 12(m + 1) + 39 = 0
m² - 12m - 12 + 39 = 0
m² - 12m + 27 = 0
D = (-12)² - 4 * 1 * 27 = 144 - 108 = 36 = 6²
![m_{1}= \frac{12+6}{2}=9\\\\ m_{2}= \frac{12-6}{2}=3\\\\\\ x^{2} +3x-1=9\\\\ x^{2} +3x-10=0\\\\D=3 ^{2} - 4 * 1 * (- 10) = 9+40=49= 7^{2}\\\\ x_{1}= \frac{-3+7}{2}=2\\\\ x_{2} = \frac{-3-7}{2}=-5 \\\\\\ x^{2} +3x-1=3\\\\ x^{2} +3x-4=0\\\\D= 3^{2}-4*1*(-4)=9+16=25= 5^{2} \\\\ x_{3}= \frac{-3+5}{2}=1 \\\\ x_{4} = \frac{-3-5}{2}=-4](https://tex.z-dn.net/?f=%20m_%7B1%7D%3D%20%5Cfrac%7B12%2B6%7D%7B2%7D%3D9%5C%5C%5C%5C%20m_%7B2%7D%3D%20%5Cfrac%7B12-6%7D%7B2%7D%3D3%5C%5C%5C%5C%5C%5C%20x%5E%7B2%7D%20%2B3x-1%3D9%5C%5C%5C%5C%20x%5E%7B2%7D%20%2B3x-10%3D0%5C%5C%5C%5CD%3D3%20%5E%7B2%7D%20%20-%204%20%2A%201%20%2A%20%28-%2010%29%20%3D%209%2B40%3D49%3D%207%5E%7B2%7D%5C%5C%5C%5C%20x_%7B1%7D%3D%20%5Cfrac%7B-3%2B7%7D%7B2%7D%3D2%5C%5C%5C%5C%20x_%7B2%7D%20%3D%20%5Cfrac%7B-3-7%7D%7B2%7D%3D-5%20%5C%5C%5C%5C%5C%5C%20x%5E%7B2%7D%20%2B3x-1%3D3%5C%5C%5C%5C%20x%5E%7B2%7D%20%2B3x-4%3D0%5C%5C%5C%5CD%3D%203%5E%7B2%7D-4%2A1%2A%28-4%29%3D9%2B16%3D25%3D%205%5E%7B2%7D%20%5C%5C%5C%5C%20x_%7B3%7D%3D%20%5Cfrac%7B-3%2B5%7D%7B2%7D%3D1%20%5C%5C%5C%5C%20x_%7B4%7D%20%3D%20%5Cfrac%7B-3-5%7D%7B2%7D%3D-4%20%20%20%20%20%20%20%20%20%20)
Х³+3х²-2=0.
Решение:тут выносить х не получится.
Разобъем 3х² на слагаемые:3х²=2х²+х².Тогда получаем
х³+3х²-2=х³+х²+2х²-2=0.Дальше группируем:
(
х³+х²)+(
2х²-2)=х²(х+1)+2(х²-1)=0
х²(х+1)+2(х²-1)=
х²(х+1)+2(х-1)(х+1)=0
(х+1)(х²+2(х-1))=0,(х+1)(х²+2х-2)=0.
тогда х₁=-1
х²+2х-2=0,D₁=(b/2)²-ac,x=-b/2+-√D₁,
D₁=1²-(-2)=3,√D₁=√3
x₂=-1+√3,x₃=-1-√3.
Ответ:
-1-√3, -1,-1+√3.