Hoặc
Tính:
a) (1−xx+x2−1):x−1x;
a, (1−xx+x2−1):x−1x
=[1−xx+x(x2−1)x]:x−1x
=−(x−1)+x(x+1)(x−1)x.xx−1
=(x−1)[−1+x(x+1)]x.xx−1
=(x−1)(−1+x2+x).xx.(x−1)
= x2 + 1 – 1;