八年级数学下分式1/(x+1)-1/(x+2)-1/(x+3)+1/(x+4)
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八年级数学下分式1/(x+1)-1/(x+2)-1/(x+3)+1/(x+4)
八年级数学下分式1/(x+1)-1/(x+2)-1/(x+3)+1/(x+4)
八年级数学下分式1/(x+1)-1/(x+2)-1/(x+3)+1/(x+4)
1/(x+1)-1/(x+2)-1/(x+3)+1/(x+4)
=1/[(x+2)(x+1)]-1/[(x+3)(x+4)]
=(4x+10)/[(x+1)(x+2)(x+3)(x+4)]
两两合并
原式=[(x+2)-(x+1)]/[(x+1)(x+2)]-[(x+4)-(x+3)]/[(x+3)(x+4)]
=1/[(x+1)(x+2)]-1/[(x+3)(x+4)]
=[(x+3)(x+4)-(x+1)(x+2)] / [(x+1)(x+2)(x+3)(x+4)]
=(4x+10)/[(x+1)(x+2)...
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两两合并
原式=[(x+2)-(x+1)]/[(x+1)(x+2)]-[(x+4)-(x+3)]/[(x+3)(x+4)]
=1/[(x+1)(x+2)]-1/[(x+3)(x+4)]
=[(x+3)(x+4)-(x+1)(x+2)] / [(x+1)(x+2)(x+3)(x+4)]
=(4x+10)/[(x+1)(x+2)(x+3)(x+4)]
你给的条件算到这,比较合适~~再往下分就繁琐了~~分子还能提取个2~~你自己看~
收起
两两合并
原式=[(x+4)+(x+1)]/[(x+1)(x+4)]-[(x+3)+(x+2)]/[(x+2)(x+3)]
=(2x+5)*{1/[(x+1)(x+4)]-1/[(x+2)(x+3)]}
=(2x+5)*{[(x+2)(x+3)-(x+1)(x+4)] / [(x+1)(x+2)(x+3)(x+4)]}
=2(2x+5)/[(x+1)(x+2)(x+3)(x+4)]
答案是o。
{1/(x+1)+1/(x+4)} -{1/(x+2)+1/(x+3)}通分,求出分子为2X+5-2x-5。
分子为0,分式结果为0