作业帮已知(3x-22)/(x+2)(x-5)=a/(x+2)-b/(x-5),则a+b=_
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已知(3x-22)/(x+2)(x-5)=a/(x+2)-b/(x-5),则a+b=_
已知(3x-22)/(x+2)(x-5)=a/(x+2)-b/(x-5),则a+b=_
已知(3x-22)/(x+2)(x-5)=a/(x+2)-b/(x-5),则a+b=_
(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
a+b=5
楼上符号貌似不对哦~
把右边通分即可
右=[a(x-5)+b(x+2)]/(x+2)(x-5)=[(a+b)x+(-5a+2b)]/(x+2)(x+5)
左右相等，所以对应项的系数相等
所以a+b=3
有不明白的再问我好了
(3x-22)/(x+2)(x-5)=a(x-5)-b(x+2)/(x+2)(x-5) =(ax-5a-bx-2b)/(x+2)(x-5)
(3x-22)/(x+2)(x-5)=[(a-b)x-(5a+2b)]/(x+2)(x-5) (3x-22)/(x+2)(x-5)=(ax-5a-bx-2b)/(x+2)(x-5)
(3x-22)与 [(a-b)x-(5a+2b)]
a-b=3 5a+2b=22
a=4 b=1
a+b=5
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