请教一道初二数学题，(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+5)-(x+8)/(x+7)
2个回答
展开全部
(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+5)-(x+8)/(x+7)
1-1/(x+1)-1-1/(x+3)=1+1/(x+5)-1-1/(x+7)
1/(x+1)-1/(x+3)=1/(x+5)-1/(x+7)
2/((x+1)(x+3))=2/((x+5)(x+7))
(x+1)(x+3)=(x+5)(x+7)
x^2+4x+3=x^2+12x+35
8x=-32
x=-4
1-1/(x+1)-1-1/(x+3)=1+1/(x+5)-1-1/(x+7)
1/(x+1)-1/(x+3)=1/(x+5)-1/(x+7)
2/((x+1)(x+3))=2/((x+5)(x+7))
(x+1)(x+3)=(x+5)(x+7)
x^2+4x+3=x^2+12x+35
8x=-32
x=-4
本回答由提问者推荐
已赞过
已踩过<
你对这个回答的评价是？
展开全部
(x+2)/(x+1)-(x+4)/(x+3)=(x+6)/(x+5)-(x+8)/(x+7)
[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)]=[(x+6)(x+7)-(x+5)(x+8)]/[(x+5)(x+7)]
2/(x^2+4x+3)=2/(x^2+12x+35)
x^2+4x+3=x^2+12x+35
4x+3=12x+35
8x=-32
x=-4
[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)]=[(x+6)(x+7)-(x+5)(x+8)]/[(x+5)(x+7)]
2/(x^2+4x+3)=2/(x^2+12x+35)
x^2+4x+3=x^2+12x+35
4x+3=12x+35
8x=-32
x=-4
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
