1/a+1/b= 4/a+b,则分式b/a+a/b
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![1/a+1/b= 4/a+b,则分式b/a+a/b](/uploads/image/z/2583844-52-4.jpg?t=1%2Fa%2B1%2Fb%3D+4%2Fa%2Bb%2C%E5%88%99%E5%88%86%E5%BC%8Fb%2Fa%2Ba%2Fb)
1/a+1/b= 4/a+b,则分式b/a+a/b
1/a+1/b= 4/a+b,则分式b/a+a/b
1/a+1/b= 4/a+b,则分式b/a+a/b
两边同时乘以ab(a+b)
b(a+b)+a(a+b)=4ab
化简:a^2-2ab+b^2=0
(a-b)^2=0
a=b
b/a+a/b=2
显然a、b都不为零
(a+b) / ab = 4 / (a+b)
(a+b)^2 = 4ab
a^2 + 2ab +b^2 = 4ab
a^2 - 2ab +b^2 = 0
(a - b)^2 = 0
a= b
分式b/a+a/b = 2