( x2 - 1)2 - x( x2 - 1) - 2x2 = 0 ( 1 )
Đặt : x2 - 1 = a , ta có :
( 1) ⇔ a2 - ax - 2x2 = 0
⇔ a2 + ax - 2ax - 2x2 = 0
⇔ a( a + x) -2x( a + x) = 0
⇔ ( a + x)( a - 2x ) = 0
TH1 : Với : a + x = 0
⇔ x2 + x - 1 = 0
⇔ x2 +\(2.\dfrac{1}{2}x+\dfrac{1}{4}-1-\dfrac{1}{4}=0\)
⇔ \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\) = 0
⇔ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
* ) \(x+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\)
⇔ \(x=\dfrac{\sqrt{5}-1}{2}\)
*) \(x+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\)
⇔\(x=-\dfrac{\sqrt{5}+1}{2}\)
TH2 . a - 2x = 0
⇔ x2 - 2x - 1 = 0
⇔ x2 - 2x + 1 - 2 = 0
⇔ ( x - 1)2 = 2
*) x - 1 = \(\sqrt{2}\)
⇔ x = \(\sqrt{2}\) + 1
*) x - 1 = - \(\sqrt{2}\)
⇔ x = 1 - \(\sqrt{2}\)
KL.....
p/s : Mk nghĩ zậy 
