Question

A/(x-5)+B/(x+2)=(5x-4)/(x的平方-3x-10)求A,B

Answer 1

左边通分=[A(x+2)+B(x-5)]/(x-5)(x+2)
=[(A+B)x+(2A-5B)]/(x-5)(x+2)
=(5x-4)/(x-5)(x+2)
则(A+B)x+(2A-5B)=5x-4
所以A+B=5
2A-5B=-4
A=3
B=2

Answer 2

A/(X-5)+B/(X+2)=(5X-4)/(X^2-3X-10)
[A(X+2)+B(X-5)]/[(X-5)(X+2)]=(5X-4)/(X^2-3X-10)
[(A+B)x+(2A-5B)]/(X^2-3X-10)=(5X-4)/(X^2-3X-10)
A+B=5 （1）
2A-5B=-4（2）
解（1） （2）得
A=3 B=2

Answer 3

A/(x-5)+B/(x+2)=(5x-4)/(x的平方-3x-10)
通分
A+B=5
2A-5B=-4
A=3,B=2

Answer 4

等号前面的式子通分后是（A+B）/x+（2A-5B）他和等号后面的（5
x-4）是对应的 列方程组得到A+B=5,2A-5B=-4 解得B=2，A=3

Answer 5

A/(x-5)+B/(x+2)=(5x-4)/(x^2-3x-10)
A/(x-5)+B/(x+2)=(5x-4)/(x-5)(x+2)
两边乘(x-5)(x+2)。A(x+2)+B(x-5)=5x-4
(A+B-5)x+2A-5B+4=0..
如果x是任意数的话，有A+B-5=0 . 2A-5B+4=0, A=3,B=2

Answer 6

解：A/(x-5)+B/(x+2)
=A(x+2)/(x-5)(x+2)+B(x+2)/(x-5)(x+2)
=(Ax+2A+Bx-5B)/(x²-3x-10)
=[(A+B)x+2A-5B]/(x²-3x-10)
∴{A+B=5
2A-5B=-4
解方程组得：{A=3
B=2