... 2 = (x 2 -x+5) 2 ⇔(x 2 +2x-5) 2 -( x 2 -x+5) 2 = 0 ⇔ ((x 2 +2x-5)+(x 2 -x+5)). ((x 2 +2x-5 )-( x 2 -x+5)) = 0 ⇔ (2x 2 + x)(3x-10) = 0 ⇔ 2x 2 +x = 0 3x-10 = 0 ⇔ x(2x+1)=0 3x-10 = 0 ... 3x 2 - 2x - 6 = 0 ⇔ x 2 (x + 3) -2 (x + 3) = 0 ⇔ (x + 3)(x 2 - 2) = 0 ⇔ x +3 = 0 x 2 -2 = 0 x 1 = -3 ⇔ x 1 = x 1 = - 2 2 Vậy phương trình có ba nghiệm: x = -3 , ,- 2 2 b, (x 2 +2x-5) ... ≠ -1 , x ≠ 4 (*) ⇔ 2 2 ( 4) 8 ( 1)( 4) ( 1)( 4) x x x x x x x x − − + = + − + − ⇔ 2x 2 - 8x = x 2 - x + 8 ⇔ x 2 - 7x - 8 = 0 ∆ = 7 2 - 4.1. (-8 ) = 81 > 0 x 1 = 7 81 2 + x 2 = = 8 = -1 ...