8x+9除以x的平方+x-6=A除以x+3加上B除以x-2

高手就做这道题

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(8x+9)/(x^2+x-6)=a/(x+3) + b/(x-2)

8x+9=a(x-2)+b(x+3)
8x+9=(a+b)x+(3b-2a)
建立握指方程组
a+b=8
3b-2a=9
得到判皮段解掘誉
a=3
b=5

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8x+9除以x的平方+x-6=A除以x+3加上配仿首B除以x-2
(8x+9)/(x^2+x-6)=A/(x+3)+B/(x-2)=[A(x-2)+B(x+3)]/大腊(x^2+x-6)
=[(A+B)x+3B-2A]/(x^2+x-6)
所以培数A+B=8
3B-2A=9
2A+2B=16
5B=25
B=5
A=8-B=3

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(8x+9)/(x^2+x-6)=A/(x+3)+B/清猛(x-2)
(8x+9)/(x+3)(x-2)=[A(x-2)+B(x+3)]/歼弯(x+3)(x-2)
8x+9=A(x-2)+B(x+3)=(A+B)x+(3B-2A)
所以答改桥A+B=8
3B-2A=9
2A+2B=16
5B=25
B=5
A=8-B=3

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