已知a,b满足a/b-b/a=2,求代数式(a^2+ab+b^2)/(b^2+4ab-a^2)的值

已知a,b满足a/b-b/a=2,求代数式(a^2+ab+b^2)/(b^2+4ab-a^2)的值。初一下册题,要过程... 已知a,b满足a/b-b/a=2,求代数式 (a^2+ab+b^2)/(b^2+4ab-a^2)的值。初一下册题,要过程 展开
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a/b-b/a=2
(a/b)^2+(b/a)^2-2=2
(a/b)^2+(b/a)^2+2=6
(a/b+b/a)^2=6
a/b+b/a=±√6

(a^2+ab+b^2)/(b^2+4ab-a^2) 分子分母同除以ab
=(a/b+1+b/a)/(b/a+4-a/b) 代入 a/b+b/a=±√6 a/b-b/a=2
=(±√6+1)/(4-2)
=(1±√6)/2
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a/b-b/a=2得a^2=b^2+2ab带入代数式得b/a+3/2
设b/a=x则由a/b-b/a=2得1/x-x=2解方程得x=根号2-1或负(根号2+1)
带入结果得 根号2+1/2 或 1/2-根号2
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