a , A = x 2 − 2 x + 1 x 2 − 1 = ( x − 1 ) 2 ( x − 1 ) ( x + 1 ) = x − 1 x + 1 a,A= x 2 −1 x 2 −2x+1 = (x−1)(x+1) (x−1) 2 = x+1 x−1 b , b, Khi x = 3 x=3 thì : x − 1 x + 1 = 3 − 1 3 + 1 = 2 4 = 1 2 x+1 x−1 = 3+1 3−1 = 4 2 = 2 1 Khi x = − 3 / 2 x=−3/2 thì : − 3 2 − 1 − 3 2 + 1 = − 3 2 − 2 2 − 3 2 + 2 2 = − 5 2 − 1 2 = − 5 2 ⋅ ( − 2 ) = 10 2 = 5 − 2 3 +1 − 2 3 −1 = − 2 3 + 2 2 − 2 3 − 2 2 = − 2 1 − 2 5 =− 2 5 ⋅(−2)= 2 10 =5 c , c, Để A A nhận giá trị nguyên ta có : x − 1 x + 1 = x + 1 − 2 x + 1 = x + 1 x + 1 − 2 x + 1 x+1 x−1 = x+1 x+1−2 = x+1 x+1 − x+1 2 Vậy x + 1 ∈ Ư ( 2 ) = { ± 1 ; ± 2 } x+1∈Ư(2)={±1;±2} − > x + 1 = 1 = > x = 0 −>x+1=1=>x=0 − > x + 1 = − 1 = > x = − 2 −>x+1=−1=>x=−2 − > x + 1 = 2 = > x = 1 −>x+1=2=>x=1 − > x + 1 = − 2 = > x = − 3 −>x+1=−2=>x=−3

Mà x khác ±1

Suy ra x thuộc{ 0;-2;-3}