Question

Решите уравнение (х+1)^4+(х+1)^2-6=0заранее спасибо)

Answer 1

( x + 1 )^4 + ( x + 1 )^2 - 6 = 0
( x + 1 )^2 = a ; a > 0
a^2 + a - 6 = 0
D = 1 + 24 = 25 = 5^2
a1 = ( - 1 + 5 ) : 2 = 2
a2 = ( - 1 - 5 ) : 2 = - 3 ( < 0 )
( x + 1 )^2 = 2
X^2 + 2x + 1 - 2 = 0
X^2 + 2x - 1 = 0
D = 4 + 4 = 8 = ( 2 V 2 )^2
X1 = ( - 2 + 2 V 2 ) : 2 = - 1 + V 2
X2 = ( - 2 - 2 V 2 ) : 2 = - 1 - V 2

Answer 2

Пусть (х+1)^2=t
t^2+t-6=0
D=1+24=25
t1=-1+5/2=2
t2=-1-5/2=-3 не подходит
(х+1)^2=2
х+1=\/2 или х+1=-\/2
Ответ: х=\/2-1; х=-\/2-1