Решите уравнение пжжжжж.

(x+3)⁴+2(x+3)²-8=0
(x+3)²=t
t≥0
t²+2t-8=0
t₁+t₂= -2 (теорема Виета)
t₁t₂= -8
t₁= -4<0 <br>t₂=2

t= 2

(x+3)²=t

(x+3)²=2
x²+6x+9=2
x²+6x+7=0
D=6²-4*7=37-28=8
$x_1= \frac{-6+ \sqrt{8} }{2} = \frac{-6+ \sqrt{4*2} }{2} = \frac{-6+ 2\sqrt{2} }{2} = \frac{2(-3+ \sqrt{2}) }{2} = \sqrt{2} -3 \\ x_2= \frac{-6- \sqrt{8} }{2} = \frac{-6- \sqrt{4*2} }{2} = \frac{-6- 2\sqrt{2} }{2} = \frac{2(-3- \sqrt{2}) }{2} =-3 -\sqrt{2} \\$