数学之美团为你解答
原式 = (x-1) / [(x+1)(x+2)] + 6/ [ (x+1)(2 - x) ] - (10-x) / (2+x)(2-x)
= (x-1)(x-2) / [(x+1)(x+2)(x-2)] - 6(x+2)/ [ (x+1)(x-2)(x+2) ] - (x -10)(x+1) / [ (x+2)(x-2)(x+1) ]
= [ (x² - 3x+2) - (6x+12) - (x² - 9x - 10) ] / [ (x+2)(x-2)(x+1) ]
= 0
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