\(x^4+2009x^2+2008x+2009\)

\(=\left(x^4+x^3+x^2\right)+\left(-x^3-x^2-x\right)+\left(2009x^2+2009x+2009\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)

Ta có:

\(x^4+2009x^2+2008x+2009\)

\(=\left(x^4+x^3+x^2\right)+\left(-x^3-x^2-x\right)+\left(2009x^2+2009x+2009\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)

x⁴+2009x²+2008x+2009

=(x⁴-x)+(2009x²+2009x+2009)

=x(x³-1)+2009(x²+x+1)

=x(x-1)(x²+x+1)+2009(x²+x+1)

=(x²+x+1)(x(x-1)+2009)

=(x²+x+1)(x²-x+2009)

k mình nha, chúc bạn học giỏi!!!

cách 1 dùng hệ số bất định
có hệ
a+c=0
ac+b+d= 2009
ad+bc=2008
bd=2009
Ta tìm được a=1,b=1,d=2009,c=-1
=> (x^2+x+1)(x^2-x+2009)=0
Cách 2:
có (x^2+m)^2 =2mx^2+m^2 +2009x^2+2009x+2009=x^2(2009+2m) +2008x +2009+m^2
xét  \delta thấy vô nghiệm => PT vô nghiệm

x4 +2009x2 +2008x + 2009 = (x4 +x3 +x2 ) + (– x3 – x2 – x) +(2009 x2 + 2009x + 2009 ) = 
x2(x2 +x + 1) – x (x2 +x + 1) + 2009 (x2 +x + 1) = (x2 +x + 1)( x2 – x + 2009)
Để ý rằng: Tam thức x2 +x + 1 có ∆ = 12 – 4 = – 3 < 0 và tam thức x2 – x + 2009 có ∆ = 12 – 4.2009 = –8035 < 0 nên các tam thức đó bất khả qui trên R . Vậy kết quả phân tích trên là kết quả cuối cùng.

 Vậy phương trình có một nghiệm x = –15.

2x^4 hay x^4

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

  x^4 + 2008x^2 + 2007x + 2008

\(=x^4+x^2+2007x^2+2007x+2007+1\)

\(=x^4+x^2+1+2007\left(x^2+x+1\right)\)

\(=\left(x^2+1\right)^2-x^2+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

x^4+2008x^2+2007x+2008

=x^4+2008x^2+2008x-x+2008

=(x^4-x)+(2008x^2+2008x+2008)

=x(x^3-1)+2008(x^2+x+1)

=x(x-1)(x^2+x+1)+2008(x^2+x+1)

=(x^2+x+1)(x^2-x+2008)

       x⁴+2008x²+2007x+2008

<=> x⁴-x+2008x²+2008x+2008

<=> x(x³-1)+2008(x²+x+1)

<=> x(x-1)(x²+x+1)+2008(x²+x+1)

<=> (x²+x+1)(x²-x+2008)

\(\left(x^4+x^2+1\right)+\left(2007x^2+2007x+2007\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2007\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2008\right)\)

x^4_x+2008(x²+x+1)=x(x-1)(x2+x+1)+2008(x2+x+1)=(x2-x+2008)(x2+x+1)

x⁴+4 = (x²)²+2² = x⁴ + 2.x².2 + 4 – 4x²

= (x² + 2)² – (2x)² = (x²-2x+2)(x²+2x+2)

Ta có: \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)