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已知x2+y2﹣8x﹣10y﹢41=0,求y/x2-xy+x/y2-xy
人气:378 ℃ 时间:2020-06-12 03:08:40
解答
x²+y²-8x-10y+41=0x²-8x+16+y²-10y+25=0(x-4)²+(y-5)²=0平方项恒非负,两平方项之和=0,两平方项均=0x-4=0 x=4y-5=0 y=5y/(x²-xy)+x/(y²-xy)=y/[x(x-y)]-x/[y(x-y)]=(y²...
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