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例1、计算(x-1)/(x~2-3x+2)+(x+1)/(x-2)-(x~2-x-6)/(x~2-4) 解:原式=(x-1)/[(x-1)(x-2)]+(x+1)/(x-2)[(x-3)(x+2)]/[(x+2)(x-2)]=1/(x-2)+(x+1)/(x-2)-(x-3)/(x-2)=[1+(x+1)-(x-3)]/(x-2)=5/(x-2) 说明:本题看起来是异分母的分式相加减,但把两个较复杂的公式的分子、分母分解因式后,约去公因式,就变简单了,且是同分母的分式相加减。若不这样做,则会异常繁杂。
Example 1. Calculation of (x-1)/(x~2-3x+2)+(x+1)/(x-2)-(x~2-x-6)/(x~2-4) solutions : Original formula = (x-1)/[(x-1)(x-2)]+(x+1)/(x-2)[(x-3)(x+2)]/[(x +2)(x-2)]=1/(x-2)+(x+1)/(x-2)-(x-3)/(x-2)=[1+(x+1) -(x - 3)]/(x - 2) = 5/(x - 2) Note: This question appears to be the addition and subtraction of the fractions of the different denominator, but the numerator and denominator of the two more complex formulae are decomposed. After the formula, about the common factor, it becomes simple, and it is a fraction with the denominator. If you do not do this, it will be very complicated.