请问这道数学题怎么做:(x+2/x+1)-(x+3/x+2)-(x-4/x-3)+(x-5/x-4)
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我给你一种好的解法:
解:分子都比分母大1,所以先整理为:
原式=(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=[1+1/(x+1)]-[1+1/(x+2)]-[1-1/(x-3)]+[1-1/(x-4)]
=1+1/(x+1)-1-1/(x+2)-1+1/(x-3)+1-1/(x-4)
=[1/(x+1)+1/(x-3)]-[1/(x+2)+1/(x-4)]
=[(x-3)+(x+1)]/[(x+1)(x-3)]-[(x-4)+(x+2)]/[(x+2)(x-4)]
=(2x-2)/(x^2-2x-3)-(2x-2)/(x^2-2x-8)
=(2x-2)*[1/(x^2-2x-3)-1/(x^2-2x-8)]
=(2x-2)*[(x^2-2x-8)-(x^2-2x-3)]/[(x^2-2x-3)(x^2-2x-8)]
=(2x-2)*(-5)/[(x^2-2x-3)(x^2-2x-8)]
=(10-10x)/[(x^2-2x-3)(x^2-2x-8)]
注:为了让你看得更清楚,我一步一步写下来,但在实际书写时,有些步骤可以省略。
解:分子都比分母大1,所以先整理为:
原式=(x+2)/(x+1)-(x+3)/(x+2)-(x-4)/(x-3)+(x-5)/(x-4)
=[1+1/(x+1)]-[1+1/(x+2)]-[1-1/(x-3)]+[1-1/(x-4)]
=1+1/(x+1)-1-1/(x+2)-1+1/(x-3)+1-1/(x-4)
=[1/(x+1)+1/(x-3)]-[1/(x+2)+1/(x-4)]
=[(x-3)+(x+1)]/[(x+1)(x-3)]-[(x-4)+(x+2)]/[(x+2)(x-4)]
=(2x-2)/(x^2-2x-3)-(2x-2)/(x^2-2x-8)
=(2x-2)*[1/(x^2-2x-3)-1/(x^2-2x-8)]
=(2x-2)*[(x^2-2x-8)-(x^2-2x-3)]/[(x^2-2x-3)(x^2-2x-8)]
=(2x-2)*(-5)/[(x^2-2x-3)(x^2-2x-8)]
=(10-10x)/[(x^2-2x-3)(x^2-2x-8)]
注:为了让你看得更清楚,我一步一步写下来,但在实际书写时,有些步骤可以省略。
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