Question

Помогите, пожалуйста, один маленький номерок решииить:3

Answer 1

x^2+2x+8=t
sqrt)t)=20-t
t=400+t^2-40t
t^2-41t+400=0
t1=25
t2=16
x^2+2x-8=0  x1=-4 x2=2
x^2+2x-17=0
x=-1+3sqrt(2)
x=-1-3sqrt(2)

Answer 2

$x^2+2x+sqrt{x^2+2x+8}=12\t=x^2+2x\\t+ sqrt{t+8}=12 \ sqrt{t+8}=12-t\\ left { {{12-t geq 0} atop {t+8=(12-t)^2}} right. \\ left { {{t leq 12} atop {t+8=144-24t+t^2}} right. \\ left { {{t leq 12} atop {t^2-25t+136=0}} right. \\D=81\t_{1}=17, t_{2}=8$

$x^2+2x=17\x^2+2x-17=0\D=2^2-4*1*(-17)=72\ x_{1}= frac{-2+ sqrt{72} }{2}= frac{-2+6 sqrt{2} }{2}= -1+3 sqrt{2}\x_{2}=-1-3 sqrt{2} \\x^2+2x=8\x^2+2x-8=0\x_{3}=2\x_{4}=-4\\\-1б3 sqrt{2};2;-4$