Question

如果x等于它的倒数，那么分式x-1分之x^2+2x-3除以x^2-3x+1分之x+3等于多少

如果x等于它的倒数，那么分式x-1分之x^2+2x-3除以x^2-3x+1分之x+3等于多少 要详细过程

Answer 1

因为x等于它的倒数,所以得:
x=1/x
x^2=1
x1=1(舍),x2=-1

(x-1)分之(x^2+2x-3)除以(x^2-3x+1)分之(x+3)
=(x-1)分之(x+3)(x-1)除以(x^2-3x+1)分之(x+3)
=(x+3)除以(x^2-3x+1)分之(x+3)
=(x+3)乘以(x+3)分之(x^2-3x+1)
=x^2-3x+1

x^2-3x+1
=1-3*(-1)+1
=1+3+1
=5

Answer 2

解：因为x等于它的倒数，
所以X=1或x=-1
由分式x-1分之x^2+2x-3除以x^2-3x+1分之x+3知，x=-1，
化简上式得x^2-3x+1,
将x=-1带入得原式=5

Answer 3

x=1/x;解x=+,-1;由题意知x=-1.(x=1时分式无意义)代入即可:

Answer 4

5，由题意x=1或x=-1