初二的几道分式题!1.已知a+b+c=0,化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) 2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)一楼的说了等于白说.........会简化的用在这问吗
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![初二的几道分式题!1.已知a+b+c=0,化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) 2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)一楼的说了等于白说.........会简化的用在这问吗](/uploads/image/z/8817372-36-2.jpg?t=%E5%88%9D%E4%BA%8C%E7%9A%84%E5%87%A0%E9%81%93%E5%88%86%E5%BC%8F%E9%A2%98%211.%E5%B7%B2%E7%9F%A5a%2Bb%2Bc%3D0%2C%E5%8C%96%E7%AE%80a%281%2Fb%2B1%2Fc%29%2Bb%281%2Fc%2B1%2Fa%29%2Bc%281%2Fa%2B1%2Fb%29+2.%E5%A6%82%E6%9E%9Ca%2Bb%2Bc%3D0%2Cabc%E4%B8%8D%E7%AD%89%E4%BA%8E0%2C%E6%B1%82bc%2F%28b%5E2%2Bc%5E2-a%5E2%29%2Bac%2F%EF%BC%88c%5E2%2Ba%5E2-b%5E2%29%2Bab%2F%28a%5E2%2Bb%5E2-c%5E2%29%E4%B8%80%E6%A5%BC%E7%9A%84%E8%AF%B4%E4%BA%86%E7%AD%89%E4%BA%8E%E7%99%BD%E8%AF%B4.........%E4%BC%9A%E7%AE%80%E5%8C%96%E7%9A%84%E7%94%A8%E5%9C%A8%E8%BF%99%E9%97%AE%E5%90%97)
初二的几道分式题!1.已知a+b+c=0,化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) 2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)一楼的说了等于白说.........会简化的用在这问吗
初二的几道分式题!
1.已知a+b+c=0,化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)
一楼的说了等于白说.........
会简化的用在这问吗
初二的几道分式题!1.已知a+b+c=0,化简a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) 2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)一楼的说了等于白说.........会简化的用在这问吗
1.已知a+b+c=0,得:
a+c=-b
a+b=-c
b+c=-a
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=a/b+a/c+b/c+b/a+c/a+c/b
=(a+c)/b+(a+b)/c+(b+c)a
=(0-b)/b+(0-c)/c+(0-a)/a
=(-1)+(-1)+(-1)
=-3
2.如果a+b+c=0,abc不等于0,求bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)
已知a+b+c=0,得:
a+c=-b
a+b=-c
b+c=-a
bc/(b^2+c^2-a^2)+ac/(c^2+a^2-b^2)+ab/(a^2+b^2-c^2)
=bc/[b^2+c^2-(b+c)^2]+ac/[c^2+a^2-(a+c)^2]+ab/[a^2+b^2-(a+b)^2)]
=bc/(-2bc)+ac/(-2ac)+ab/(-2ab)
=(-1)/2+(-1)/2+(-1)/2
=-1.5
1、原式=(b+c)/a+(a+c)/b+(a+b)/c,把c=-(a+b)代入化简得原式=-1-1-1=-3 2、同理把c=-(a+b)代入化简会发现每个分式都为-1/2,所以原式=-3/2
简单
化简一下就能做了啊