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x y为实数 且x²-2x+√xy-2=-1 求1/xy+1/﹙x+1﹚﹙y+1﹚+.+1/﹙x+2006﹚﹙y+2006﹚的值
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人气:147 ℃ 时间:2020-06-01 05:37:04
解答
(x-1)²+√(xy-2)=0
∴x-1=0
xy-2=0
∴x=1
y=2
1/xy+1/﹙x+1﹚﹙y+1﹚+.+1/﹙x+2006﹚﹙y+2006﹚
=1/1*2+1/2*3+1/3*4+……+1/2007*2008
=1-1/2+1/2-1/3+1/3-……-1/2007+1/2007-1/2008
=1-1/2008
=2007/2008(x-1)²+√(xy-2)=0怎么来的x²-2x+√xy-2=-1x²-2x+1+√xy-2=0(x-1)²+√(xy-2)=0
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