计算:(x-1)/(x^2 + 3x + 2) + 6/(2 + x - x^2) - (10-x)/(4-x^2) 答案好象是0
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(x-1)/(x^2 + 3x + 2) + 6/(2 + x - x^2) - (10-x)/(4-x^2)
=(x-1)/(x+1)(x+2)-6/(x-2)(x+1)-(10-x)/(x+2)(x-2)
=[(x-1)(x-2)-6(x+2)-(10+x)(x+1)]/(x+1)(x+2)(x-2)
=(x^2-3x+2-6x-12-10x-10-x^2-x)/(x+1)(x+2)(x-2)
=(-20-20x)/(x+1)(x+2)(x-2)
=-20(1+x)/(x+1)(x+2)(x-2)
=-20/(x+2)(x-2)
=(x-1)/(x+1)(x+2)-6/(x-2)(x+1)-(10-x)/(x+2)(x-2)
=[(x-1)(x-2)-6(x+2)-(10+x)(x+1)]/(x+1)(x+2)(x-2)
=(x^2-3x+2-6x-12-10x-10-x^2-x)/(x+1)(x+2)(x-2)
=(-20-20x)/(x+1)(x+2)(x-2)
=-20(1+x)/(x+1)(x+2)(x-2)
=-20/(x+2)(x-2)
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