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)]
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)]
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第1个回答 2019-10-04
解:(x+2)/(x+1)=1+1/(x+1)
(x+3)/(x+2)=1+1/(x+2)
利用这个思路同样可以有
(x-4)/(x-3)=1-1/(x-3)
(x-5)/(x-4)=1-1/(x-4)
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/(x+1)-1/(x+2)+1/(x-3)-1/(x-4)
=1/[(x+1)(x+2)]-1/[(x-4)(x-3)]
=[(x-4)(x-3)-(x+1)(x+2)]/[(x+1)(x+2)(x-3)(x-4)]
=(-7x+12-3x-2)/[(x+1)(x+2)(x-3)(x-4)]
=-10(x-1)/[(x+1)(x+2)(x-3)(x-4)]
(x+3)/(x+2)=1+1/(x+2)
利用这个思路同样可以有
(x-4)/(x-3)=1-1/(x-3)
(x-5)/(x-4)=1-1/(x-4)
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/(x+1)-1/(x+2)+1/(x-3)-1/(x-4)
=1/[(x+1)(x+2)]-1/[(x-4)(x-3)]
=[(x-4)(x-3)-(x+1)(x+2)]/[(x+1)(x+2)(x-3)(x-4)]
=(-7x+12-3x-2)/[(x+1)(x+2)(x-3)(x-4)]
=-10(x-1)/[(x+1)(x+2)(x-3)(x-4)]
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