( 1 ĐIỂM) 4 3 2X +2008X −2008X +2008X 2009 0− =⇔ − + + + =(X 1)(...

Bài 2 ( 1 điểm)

4

3

2

x

+

2008x

−

2008x

+

2008x 2009 0

−

=

⇔

−

+

+ +

=

(x 1)(x

2009x

x 2009) 0

3

2

⇔

−





+

+ +





=

(x 1) x (x 2009) (x 2009)

0

2

⇔

−

+

+ =

(x 1)(x 2009)(x

1) 0



− =

x 1 0



=

x 2009 0

x 1

⇒





 + =



+

=

⇒ 



= −

x

2009

x

1 0 (VN)