已知abc均为实数,且（x²-1）（x-2）分之5-7x等于x-1分之a,+x+1分之b,+x-2分之c,则abc=?

(5-7x) / {x^2-1)(x-2) }
= a/(x-1) + b/(x+1) + c/(x-2)
= { a(x+1)(x-2) + b(x-1)(x-2) + c/(x-1)(x+1) } / {x^2-1)(x-2) }
= { a(x^2-x-2) + b(x^2-3x+2) + c(x^2-1) } / {x^2-1)(x-2) }
= { (a+b+c)x^2 + (-a-3b)x + (-2a+2b-c) } / {x^2-1)(x-2) }
所以：
a+b+c = 0
-a-3b = -7
-2a+2b-c = 5
解得：a=1,b=2,c=-3
abc=1*2*(-3) = - 6