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已知x^2+y^2-8x+12y+52=0,求1/2x-1/x-y(x-y/2x-x^2+y^2)
题目详情
已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2)
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答案和解析
即(x^2-8x+16)+(y^2+12y+36)=0
(x-4)^2+(y+6)^2=0
所以x-4=0,y+6=0
x=4,y=-6
所以原式=1/2x-1/(x-y)*[(x-y)/x-(x+y)(x-y)]
=1/2x-1/x+(x+y)
=-1/(2x)+x+y
=-1/8+4-6
=-17/8
(x-4)^2+(y+6)^2=0
所以x-4=0,y+6=0
x=4,y=-6
所以原式=1/2x-1/(x-y)*[(x-y)/x-(x+y)(x-y)]
=1/2x-1/x+(x+y)
=-1/(2x)+x+y
=-1/8+4-6
=-17/8
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