It is known that for any x, there are (x + 4) = (x-1) a + (x + 3) = (BX + C) Try to determine a, B, C
∵(x+4)/(x²+2x-3)=A/(x-1)+(Bx+C)/(x²+2x-3)
Two sides are multiplied by (X & # 178; + 2x-3)
x+4=A(x+3)+Bx+C
x+4=(A+B)x+(3A+C)
∴A+B=1
3A+C=4
∴B=1-A C=4-3A
Let a = 1, B = 0, C = 1
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