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,
问题描述：
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,
答
x=1/x
x^2=1
x=1或x=-1
[(x^2+2x-3)/(x-1)]/[(x+3)/(x^2-3x+1)]
=[(1+2x-3)/(x-1)]/[(x+3)/(1-3x+1)]
=[(2x-2)/(x-1)]/[(x+3)/(2-3x)]
=2/[(x+3)/(2-3x)]
=2/[(x+3)/(2-3x)]
x=1, 2/[(x+3)/(2-3x)]=2/[4/(-1)]=-1/2
x=-1, 2/[(x+3)/(2-3x)]=2/(2/5)=5
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