(х+1)^2 (х^2+2х)=0
(Х^2+3х+1)(х^2+3х+3)=-1
(Х^2-4х+1)(х^2-4х+2)=12
Решите пжж

Ответ учителя по предмету Алгебра

(х+1)^2 (х^2+2х)=0
x+1=0
x=-1
x²+2x=0

x(x+2)=0

x=-2 x=0

(Х^2+3х+1)(х^2+3х+3)=-1
(x²+3x+1)(x²+3x+3)+1=0

(t+1)(t+3)+1=0

(t²+3t+t+3)+1=0

(t²+4t+3)+1=0

t²+4t+3+1=0

t²+4t+4=0

D=b²-4ac=4²-4*1*4=0

t₁.₂=-b/2a=-4/2*1=-2

x²+3x=-2

x²+3x+2=0

X₁.₂-b+-√D

/2a

x₁=-3-1/2*1=-2 x₂=-3+1/2*1=-1

x=-2 x=-1

(Х^2-4х+1)(х^2-4х+2)=12
(t+1)(t+2)-12=0
(t²+2t+t+2)-12=0

(t²+3t+2)-12=0

t²+3t+2-12=0

D=b²-4ac=3²-4*1(-10)=49

t₁.₂=-b+-√D/2a

t₁=-3-7/2*1=-5 t₂=-3+7/2*1=2

x²-4x=-5 x²-4x=2

x²-4x=-5

x²-4x+5=0

x²-4x=2

x²-4x-2=0

D=b²-4ac=(-4)²=4*1(-2)=24

x₁.₂=-b+-√D/2a

x₁=4-2√6/2*1=2-√6 x₂=4+2√6/2*1=2+√6

x=2-√6 x=2+√6

ответ:x=2-√6 x=2+√6

(х^2+1)(х^2+2х)=12
(x²+1)(x²+2x)-12=0

(x^4+2x³+x²+2x)-12=0

x^4+2x³+x²+2x-12=0