Theo de ta co:
1) a.x2+b.x+c = -2 . Thay x=0 vao bieu thuc nay duoc:
a.02 + b.0 + c = -2
=> 0+0+c = -2
=> c=-2
2) a.x2 + b.x+c = 1 . Thay x=1 vao bieu thuc nay duoc:
a.12 + b.1 + c = 1
=> a+ b + c = 1
Thay c=-2 vua tim o (1) vaobieu thuc tren duoc:
a+b-2 =1 => a+b =3
2) a.x2 +b.x +c = 4 . Thay x=-2 vao bieu thuc nay, ta duoc:
a.(-2)2 + b.(-2) + c = 4
=> 4a - 2b + c = 4
=> 2 ( 2a - b ) +c = 4
Thay: c = -2 tim o (1) vao bieu thuc tren duoc:
2(2a-b) -2 = 4
=> 2a- b = (4+2):2 = 3
Bay gio ta da co 2 yeu to:
a+b = 3 ; 2a-b = 3
Tu a+b = 3 => b = 3-a . Thay b=3-a vao bieeu thuc 2a-b = 3 ta duoc:
2a - (3-a) = 3 => 2a - 3 + a = 3 => 3a = 3+3 =6 => a = 6:3 = 2
Suuy ra: b = 3-2 = 1
Vay: a=2 ; b=1 ; c = -2