((Х-1)/х)-((х+1)/х-1)<2

To solve the inequality

\[ \frac{(x-1)}{x} - \left(\frac{(x+1)}{x} - 1\right) < 2, \]

let's start by simplifying the left-hand side of the inequality.

1. The expression \(\frac{(x+1)}{x} - 1\) can be simplified:

\[ \frac{x+1}{x} - 1 = \frac{x+1 - x}{x} = \frac{1}{x}. \]

So, the entire left-hand side becomes:

\[ \frac{x-1}{x} - \frac{1}{x} = \frac{x-1-1}{x} = \frac{x-2}{x}. \]

2. Now we can rewrite the inequality:

\[ \frac{x-2}{x} < 2. \]

3. To eliminate the fraction, multiply both sides of the inequality by \(x\). Note that the sign of the inequality will change if \(x < 0\), so we will consider the cases where \(x > 0\) and \(x < 0\) separately.

For \(x > 0\):

\[ x - 2 < 2x. \]

Subtract \(x\) from both sides:

\[ -2 < x. \]

Thus, for \(x > 0\), we have \(x > 2\).

4. For \(x < 0\), the inequality becomes:

\[ x - 2 > 2x. \]

Rearranging gives:

\[ -2 > x. \]

5. Therefore, combining both cases, we find the solutions:

• For \(x > 0\), we have \(x > 2\).
• For \(x < 0\), we have \(x < -2\).

6. Finally, the solution to the inequality is:

\[ x < -2 \quad \text{or} \quad x > 2. \]

In interval notation, the solution is:

\[ (-\infty, -2) \cup (2, \infty). \]