[(x+3分之x+2)-(x+2分之x+1)+(x+5分之x+4)-(x+4分之x+3)]除以[(x平方+8x+15)分

[(x+3分之x+2)-(x+2分之x+1)+(x+5分之x+4)-(x+4分之x+3)]除以[(x平方+8x+15)分之(x平方+7x+13)]
seshawu 1年前 已收到5个回答 举报

adong1976 春芽

共回答了19个问题采纳率:89.5% 举报

原式=
[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x
+3)(x+5)](x^2+7x+13)
=[(1分之x+2)-(1分之x+3)+(1分之x+4)-(1分之x+5)]*)]*[(x
+3)(x+5)](x^2+7x+13)
=[(3分之(x+2)(x+5))-(1分之(x+3)(x+4))]*[(x
+3)(x+5)](x^2+7x+13)
=[3(x+3)](x+2)-(x+5x+4)](x^2+7x+13)
=[3(x+3)(x+4)-(x+2)(x+5)][(x+2)(x+4)](x^2+7x+13)
=[3(x^2+7x+12)-(x^2+7x+100][(x+2)(x+4)](x^2+7x+13)
=(2x^2+14x+26)[(x+2)(x+4)](x^2+7x+13)
=2[(x+2)(x+4)]
=(1分之x+2)-(1分之x+4)
请采纳我的答案,谢谢

1年前

4

学着kk 幼苗

共回答了127个问题 举报

(x+2)/(x+3)-(x+1)/(x+2)+(x+4)/(x+5)-(x+3)/(x+4)
=[1-1/(x+3)]-[1-1/(x+2)]+[1-1/(x+5)]-[1-1/(x+4)]
=1/(x+2)-1/(x+3)+1/(x+4)-1/(x+5)
=(x+3-x-2)/(x+2)(x+3)+(x+5-x-4)/(x+4)(x+5)
=1/(x^2...

1年前

2

ST11 幼苗

共回答了14个问题 举报

化简:
原式=[1-1/(x+3)-1+1/(x+2)+1-1/(x+5)-1+1/(x+4)]*(x^2+7x+13)/(x+3)(x+5)
=[1/(x+2)-1/(x+3)+1/(x+4)-1/(x+5)])]*(x^2+7x+13)/(x+3)(x+5)
=[1/(x+2)(x+3)+1/(x+4)(x+5)])]*(x^2+7x+13)/(x+3)(x+5)
=2(x^2+7x+13)/(x+2)(x+3)(x+4)(x+5))*(x^2+7x+13)/(x+3)(x+5)
=2/(x+2)(x+4)

1年前

2

lihuibkd 幼苗

共回答了2个问题 举报

原式=[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x
+3)(x+5)](x^2+7x+13)
=[(1分之x+2)-(1分之x+3)+(1分之x+4)-(1分之x+5)]*)]*[(x
+3)(x+5)](x^2+7x+13)
=[(3分之(x+2)(x+5))-(1分之(x+3)(x+4))]*[...

1年前

2

洛阳旅游 幼苗

共回答了477个问题 举报

[(x+2)/(x+3)-(x+1)/(x+2)+(x+4)/(x+5)-(x+3)/(x+4)]/[(x^2+7x+13)]/(x^2+8x+15)]
={1-1/(x+3)-[1-1/(x+2)]+1-1/(x+5)-[1-1/(x+4)]}/[(x^2+7x+13)]/(x^2+8x+15)]
=1/(x+2)-1/(x+3)+1/(x+4)-1/(x+5)/[(x^2+7x...

1年前

0
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