To solve the inequation \((|x+1|-x)/x < 1\), we will follow these steps: ### Step 1: Rewrite the Inequation We start with the given inequation: \[ \frac{|x+1|-x}{x} < 1 \] ### Step 2: Move All Terms to One Side We can rearrange the inequation by subtracting 1 from both sides: \[ \frac{|x+1|-x}{x} - 1 < 0 \] This simplifies to: \[ \frac{|x+1|-x - x}{x} < 0 \] \[ \frac{|x+1| - 2x}{x} < 0 \] ### Step 3: Analyze the Absolute Value The expression \(|x+1|\) can be split into two cases based on the value of \(x\). **Case 1:** \(x + 1 \geq 0\) (i.e., \(x \geq -1\)) In this case, \(|x+1| = x + 1\). Substituting this into the inequation gives: \[ \frac{(x + 1) - 2x}{x} < 0 \] \[ \frac{1 - x}{x} < 0 \] **Case 2:** \(x + 1 < 0\) (i.e., \(x < -1\)) In this case, \(|x+1| = -(x + 1)\). Substituting this into the inequation gives: \[ \frac{-(x + 1) - 2x}{x} < 0 \] \[ \frac{-3x - 1}{x} < 0 \] ### Step 4: Solve Each Case **For Case 1:** \[ \frac{1 - x}{x} < 0 \] This inequality holds when the numerator and denominator have opposite signs. We analyze the critical points: - \(1 - x = 0 \Rightarrow x = 1\) - \(x = 0\) We test intervals: 1. \(x < 0\) (e.g., \(x = -1\)): \(\frac{1 - (-1)}{-1} = \frac{2}{-1} < 0\) (True) 2. \(0 < x < 1\) (e.g., \(x = 0.5\)): \(\frac{1 - 0.5}{0.5} = \frac{0.5}{0.5} > 0\) (False) 3. \(x > 1\) (e.g., \(x = 2\)): \(\frac{1 - 2}{2} = \frac{-1}{2} < 0\) (True) Thus, for Case 1, the solution is: \[ x \in (-\infty, 0) \cup (1, \infty) \] **For Case 2:** \[ \frac{-3x - 1}{x} < 0 \] Again, we analyze the critical points: - \(-3x - 1 = 0 \Rightarrow x = -\frac{1}{3}\) - \(x = 0\) We test intervals: 1. \(x < -1\) (e.g., \(x = -2\)): \(\frac{-3(-2) - 1}{-2} = \frac{6 - 1}{-2} = \frac{5}{-2} < 0\) (True) 2. \(-1 < x < 0\) (e.g., \(x = -0.5\)): \(\frac{-3(-0.5) - 1}{-0.5} = \frac{1.5 - 1}{-0.5} = \frac{0.5}{-0.5} < 0\) (False) 3. \(x > 0\) (e.g., \(x = 1\)): \(\frac{-3(1) - 1}{1} = \frac{-3 - 1}{1} < 0\) (False) Thus, for Case 2, the solution is: \[ x \in (-\infty, -1) \] ### Step 5: Combine Solutions The combined solution from both cases is: \[ x \in (-\infty, -1) \cup (-\infty, 0) \cup (1, \infty) \] This simplifies to: \[ x \in (-\infty, 0) \cup (1, \infty) \] ### Final Solution The final solution to the inequation is: \[ x \in (-\infty, 0) \cup (1, \infty) \]

Solve the inequation ` (|x+1|-x)/xlt1 `

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