2个回答
展开全部
(x^1/2-y^1/2)/(x^1/2+y^1/2)
分母有理化
=(x^1/2-y^1/2)^2/(x-y)
=(x+y-2x^1/2y^1/2)/(x-y)
=(12-2*3)/(x-y)
=6/(x-y)
x-y=-[(x-y)^2]^1/2 (此处取负号,因为x〈y)
=-[(x+y)^2-4xy]^1/2
=-(12^2-4*9)^1/2
=-6根号3
所以(x^1/2-y^1/2)/(x^1/2+y^1/2)
=-1/根号3
分母有理化
=(x^1/2-y^1/2)^2/(x-y)
=(x+y-2x^1/2y^1/2)/(x-y)
=(12-2*3)/(x-y)
=6/(x-y)
x-y=-[(x-y)^2]^1/2 (此处取负号,因为x〈y)
=-[(x+y)^2-4xy]^1/2
=-(12^2-4*9)^1/2
=-6根号3
所以(x^1/2-y^1/2)/(x^1/2+y^1/2)
=-1/根号3
追问
我感觉,这个是错的,。
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