1/X^2-1>=1/X-1

Let me show you a way to solve this....    

1 / [ x^2 - 1] >= 1/ [x - 1]       and we can write

1/[x^2 - 1] - 1/[x - 1]  >= 0     and getting a common denominator, we have

-x / [ x^2 - 1] > = 0   

Note that this function  does not exist at -1 or 1 because that would make the denominator 0

And note that when x = 0, the inequality is true

So the solutions  come from one, or more, of these intervals :

(-infinity, -1), (-1, 0], [0, 1), (1, infinity)

Notice that if x < -1  the inequality is true

And if   -1 < x < 0     the inequality is false

And if   0 <= x < 1   the inequality is true

And if   x > 1, the inequality is false

Here's the graph :  https://www.desmos.com/calculator/5qbwbakrao

CPhill thank you/I unerstood everuthing thants to you,bu i have another question?is that necessary to add brackets at the denominator ?

Yep...we need the brackets because the way you have it written is interpreted as:

1/x^2   - 1   >=    1/x - 1           [ the 1's  are separate from the fractions ]

Instead of what you intended which is :

1/ [x^2 -1 ]   >=  1 /[ x - 1 ]

Brackets and parentheses are important......!!!!