2 ( x − 2 ) x 2 − 1 − 3 x + 1 = {\displaystyle {2(x-2) \over x^{2}-1}-{3 \over x+1}=}
2 x − 4 x 2 − 1 − 3 ( x − 1 ) ( x + 1 ) ( x − 1 ) = {\displaystyle {2x-4 \over x^{2}-1}-{3(x-1) \over (x+1)(x-1)}=}
2 x − 4 x 2 − 1 − ( 3 x − 3 ) ( x + 1 ) ( x − 1 ) = {\displaystyle {2x-4 \over x^{2}-1}-{(3x-3) \over (x+1)(x-1)}=}
2 x − 4 − ( 3 x − 3 ) ( x + 1 ) ( x − 1 ) = {\displaystyle {2x-4-(3x-3) \over (x+1)(x-1)}=}
2 x − 4 − 3 x + 3 ( x + 1 ) ( x − 1 ) = {\displaystyle {2x-4-3x+3 \over (x+1)(x-1)}=}
− x − 1 ( x + 1 ) ( x − 1 ) = {\displaystyle {-x-1 \over (x+1)(x-1)}=}
− ( x + 1 ) ( x + 1 ) ( x − 1 ) = {\displaystyle {-(x+1) \over (x+1)(x-1)}=}
− 1 ( x − 1 ) = {\displaystyle {-1 \over (x-1)}=}
1 1 − x {\displaystyle {1 \over 1-x}}
