一道数学计算题:先化简,再求值((1/x-y)+(1/x+y))/(2y/x^2-2xy+y^2),其中x=1+根2,y=1-根2.、

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一道数学计算题:先化简,再求值((1/x-y)+(1/x+y))/(2y/x^2-2xy+y^2),其中x=1+根2,y=1-根2.、
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一道数学计算题:先化简,再求值((1/x-y)+(1/x+y))/(2y/x^2-2xy+y^2),其中x=1+根2,y=1-根2.、
一道数学计算题:先化简,再求值
((1/x-y)+(1/x+y))/(2y/x^2-2xy+y^2),其中x=1+根2,y=1-根2.、

一道数学计算题:先化简,再求值((1/x-y)+(1/x+y))/(2y/x^2-2xy+y^2),其中x=1+根2,y=1-根2.、
[1/(x-y)+1/(x+y)]/[2y/(x^2-2xy+y^2)]
={[(x+y)+(x-y)]/[(x+y)(x-y)]}*(x^2-2xy+y^2)/2y
={2x/[(x+y)(x-y)]}*{(x-y)^2/2y}
=(x/y)*(x-y)/(x+y)
x=1+根号2,y=1-根号2,x-y=2倍根号2,x+y=2,代入
原式=[(1+根号2)/(1-根号2)]*根号2
=-3倍根2-4

[1/(x-y)-1/(x+y)]/[2y/(x^2-2xy+y^2)]
={[(x+y)-(x-y)]/[(x+y)(x-y)]}*(x^2-2xy+y^2)/2y
={2y/[(x+y)(x-y)]}*{(x-y)^2/2y}
=(x-y)/(x+y)
x=1+根号2,y=1-根号2,代入
原式=【(1+根号2)-(1-根号2)】/【(1+根号2)+(1-根号2)】
=(2根号2)/2
=根号2

对的,与我算的一样。

应该是对的,相信自己~~