[(1)/(x-y)+(1)/(x+y)]÷(x的平方y)/(x的平方-y的平方),其中x=根号3+1,y=根号3-1.
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[(1)/(x-y)+(1)/(x+y)]÷(x的平方y)/(x的平方-y的平方),其中x=根号3+1,y=根号3-1.
[(1)/(x-y)+(1)/(x+y)]÷(x的平方y)/(x的平方-y的平方),其中x=根号3+1,y=根号3-1.
[(1)/(x-y)+(1)/(x+y)]÷(x的平方y)/(x的平方-y的平方),其中x=根号3+1,y=根号3-1.
先将第一个括号中的分式通分:
[x+y/(x-y)(x+y)+x-y/(x-y)(x+y)]÷(x的平方y)/(x的平方-y的平方)
合并括号中的得:
[x+y+x-y/(x-y)(x+y)]÷(x的平方y)/(x的平方-y的平方)=[2x/(x-y)(x+y)]÷(x的平方y)/(x的平方-y的平方)
将除号后面的式子转换一下,分母利用平方差公式:
[2x/(x-y)(x+y)]÷(x的平方y)/(x-y)(x+y)
所以2x/(x-y)(x+y)×(x-y)(x+y)/(x的平方y)=2x/(x的平方y)=2/xy
最后将x,y的值代入就可以了,最后答案为 根号2
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