已知3x-4除以(x-1)(x-4)=a除以x-1+b除以x-2 求整式a b

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已知3x-4除以(x-1)(x-4)=a除以x-1+b除以x-2 求整式a b
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已知3x-4除以(x-1)(x-4)=a除以x-1+b除以x-2 求整式a b
已知3x-4除以(x-1)(x-4)=a除以x-1+b除以x-2 求整式a b

已知3x-4除以(x-1)(x-4)=a除以x-1+b除以x-2 求整式a b
题目不对

(3x-4)/[(x-1)(x-2)]=a/(x-1)+b/(x-2)
两边同时乘以(x-1)(x-2)则
3x-4=a(x-2)+b(x-1)=ax-2a+bx-b
所以a+b=3
-2a-b=-4
a=1
将a=1带入式子a+b=3得b=2