(3 ĐIỂM) − + − + + × − + − −2 2X X 1 3 3(X 1) 3(X X 1) 2X 2=+ − + + +...
Bài 1: (3 điểm) − + − + + × − + − −
2
2
x x 1 3 3(x 1) 3(x x 1) 2x 2=+ − + + + + (0.5 đ)2
(x 1)(x x 1) (x 1)(x 2) x(x 2)+ + × − + − −x 2x 1 3(x x 1) 2x 2= + − −3(x 1) 2x 2 = + + + (1.0 đ)(x 1) (x 2) x(x 2)+ =− +x 2 1= 3x 2x 2+ = + (1.0 đ) x(x 2)x(x 2) xBài 2: (4 điểm) a) ĐK: x – 1 ≠ 0 ⇔ x ≠ 1. (0.5 đ)− + −3
2
x 2x 4x 1b) A = − = x2
– x + 3 + 2x 1− . (1.0 đ)x 1Với x ∈Z thì x2
–x + 3 ∈Z ⇒ A ∈ Z ⇔ 2x 1− ∈Z (0.5 đ) ⇔ x – 1 ∈ Ư(2) ⇔ x – 1 ∈