若实数x,y,z满足方程组 xy/﹙x+2y﹚=1① yz/﹙y+2z﹚=2② zx/﹙z+2x﹚③ 则
若实数x,y,z满足方程组 xy/﹙x+2y﹚=1① yz/﹙y+2z﹚=2② zx/﹙z+2x﹚③ 则
A.x+2y+3z=0 B.7x+5y+2z=0 C.9x+6y+sz=0 D.10x+7y+z=0
zx/﹙z+2x﹚=3③
人气:311 ℃ 时间:2019-12-12 08:52:12
解答
对等式①,②,③取倒数可得2/x + 1/y = 1 (1)2/y + 1/z = 1/2 (2)2/z + 1/x = 1/3 (3)(1)* 4 - (2)*2 + (3)可得9/x = 10/3,x=27/10,把x=27/10带入(1)可得y= 27/7,把x=27/10带入(3)可得z= -54.所以10x+7y+z=0,答案...
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