91Ó°ÊÓ

We start by breaking this into two cases depending on the sign of \(x-1\) and \(x+1\). Case 1: If \(x-1 \geq 0\) and \(x+1 \geq 0\), then the inequality becomes \(x-1 > x+1\), which solves to \(x < 1\). Case 2: If \(x-1 < 0\) and \(x+1 \leq 0\), the inequality becomes \(-(x-1) > -(x+1)\), which solves to \(x > -1\). Considering both cases, it appears there are no values of \(x\) that satisfy both conditions simultaneously.

Similar to step 1, this will be broken down based on the sign of \(x\) and \(x+1\). Case 1: If \(x \geq 0\) and \(x+1 \geq 0\), the inequality becomes \(x + x+1 < 2\), which solves to \(x < 0.5\). Case 2: For \(x \geq 0\) and \(x+1 < 0\), the inequality becomes \(x - (x+1) < 2\), which yields no solution as it implies \(x < -1\) (but we're considering \(x \geq 0\) in this case). Case 3: For \(x < 0\) and \(x+1 \geq 0\), the inequality becomes \(-(x + x+1) < 2\), which again yields no solution for similar reasons to Case 2. Case 4: For \(x < 0\) and \(x+1 < 0\), the inequality becomes \(-(x) - (x+1) < 2\), which solves to \(x > -1\). However, this condition doesn't match Case 4's assumption that \(x < 0\). Thus, the only solution here is when \(x\) satisfy Case 1, i.e. \(x < 0.5\).