Question

шестое! помогите! методом заміни змінної

Answer 1

3(x² + 5x + 1)² + 2x² + 10x = 3

3(x² + 5x + 1)² + 2(x² + 5x + 1) - 5 = 0

x² + 5x + 1 = m

3m² + 2m - 5 = 0

D = 2² - 4 * 3 * (- 5) = 4 + 60 = 64 = 8²

$m_{1}=\frac{-2-8}{6}=-1\frac{2}{3}\\\\m_{2}=\frac{-2+8}{6}=1\\\\1)x^{2}+5x+1=-1\frac{2}{3}\\\\3x^{2}+15x+3=-5\\\\3x^{2}+15x+8=0\\\\D=15^{2}-4*3*8=225-96=129\\\\x_{1}=\frac{-15-\sqrt{129}}{6}\\\\x_{2}=\frac{-15+\sqrt{129} }{6}$

$2)x^{2}+5x+1=1\\\\x^{2}+5x=0\\\\x(x+5)=0\\\\x_{3} =0\\\\x+5=0\\\\x_{4}=-5$

Answer 2

Решение смотрите во вложении