Question

giải PT:  x⁴+3x³+4x²+3x+1=0

Answer 1

đặt P(x)=x^4+3x^3+4x^2+3x+1

đặt y=x²+1

=>y²=(x²+1)²

=>y²=x⁴+2x²+1

=>P(x)=x⁴+2x²+1+3x³+2x²+3x

=x⁴+2x²+1+3x³+3x+2x²

=x⁴+2x²+1+3x(x²+1)+2x²

=y²+3xy+2x²

=y²+xy+2xy+2x²

=y(y+x)+2x(y+x)

=(y+x)(y+2x)

thay y=x²+1 ta được:

P(x)=(x²+1+x)(x²+1+2x)

=>x^4+3x^3+4x^2+3x+1=0

<=>(x²+1+x)(x²+1+2x)=0

<=>x²+1+x=0 hoặc x²+1+2x=0

mà x²\(\ge\)|x|

nên x²+x\(\ge\)0 

=>x²+1+x>0

nên x²+1+2x=0

<=>(x+1)²=0

<=>x+1=0

<=>x=-1