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若x等于它的倒数,求x-1分之x的平方+2x-3除以x的平方-3x+1分之x+3,
人气:145 ℃ 时间:2019-08-20 12:06:56
解答
x=1/x,x²=1,x=±1;
[(x²+2x-3)/(x-1)]/[(x+3)/(x²-3x+1)]
=(x+3)(x-1)(x²-3x+1)/[(x-1)(x+3)]
=x²-3x+1
=1±1*3+1=2±3
=5或-1
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