Question

已知y=x²-2x+1/x²-1÷x²-x/x+1-1/x+1是说明在等式右边有意义的条件

Answer 1

y=[(x^2-2x+1)/(x^2-1)]/[(x^2-x)/(x+1)]-1/x+1

=[(x-1)^2/(x+1)(x-1)]/[x(x-1)/(x+1)]-1/x+1

=[(x-1)/(x+1)]*[(x+1)/x(x-1)]-1/x+1

=1/x-1/x+1

=1

所以无论x取何值,Y的值不变