方法1、同分系数求值(3x-22)/(x+2)(x-5)=a(x-5)-b(x+2)/(x+2)(x-5)————右边通分
=(ax-5a-bx-2b)/(x+2)(x-5)
=[(a-b)x-(5a+2b)]/(x+2)(x-5)
(3x-22)与 [(a-b)x-(5a+2b)]系数相等
a-b=3 5a+2b=22
a=4 b=1
方法2、等式的基本性质2：
3x-22/(x+2)·(x-5)=(a/x+2)-(b/x-5),
两边同乘以x+2,得
3x-22/(x-5)=a-b/(x-5)· (x+2)
令x=-2,得
a=4
两边同乘以x-5,得
3x-22/(x+2)=a/(x+2)·(x-5)-b
令x=5,得
-b=-7/7=-1
b=1
即a=4,b=1