<p>Solve <br>\(\left|\dfrac{x+1}{x}\right| + |x+1| = \dfrac{(x+1)^2}{|x|}\)</p>
<p>\(x \in (0, \infty) \cup \{-1\}\)</p>
<p>\(x \in (-\infty, 0)\)</p>
<p>\(x \in \{-1\}\)</p>
<p>\(x \in (0, 1)\)</p>
Step-by-Step Solution
Key Concept: Recognize that the RHS equals |x+1|·|x+1|/|x|, allowing you to factor the equation as |x+1|·(|x+1|/|x| - 1 - |x|) = 0. This transforms an intimidating absolute value equation into manageable cases.
<p><strong>Step 1:</strong> Note that the domain requires x ≠ 0.</p><p><strong>Step 2:</strong> Rewrite the equation: |x+1|/|x| + |x+1| = (x+1)²/|x|</p><p><strong>Step 3:</strong> Recognize that (x+1)²/|x| = |x+1|·|x+1|/|x| = |x+1|·(|x+1|/|x|)</p><p><strong>Step 4:</strong> Factor out |x+1|: |x+1|·(|x+1|/|x| - 1) = |x+1|</p><p><strong>Step 5:</strong> This gives |x+1|·(|x+1|/|x| - |x|/|x|) = |x+1|, or |x+1|·[(|x+1| - |x|)/|x|] = |x+1|</p><p><strong>Step 6:</strong> Either |x+1| = 0 (giving x = -1) or (|x+1| - |x|)/|x| = 1, meaning |x+1| - |x| = |x|, so |x+1| = 2|x|</p><p><strong>Step 7:</strong> For |x+1| = 2|x|: Square both sides: (x+1)² = 4x², giving x² + 2x + 1 = 4x², so 3x² - 2x - 1 = 0</p><p><strong>Step 8:</strong> Factoring: (3x + 1)(x - 1) = 0, giving x = 1 or x = -1/3</p><p><strong>Step 9:</strong> Verify all solutions in the original equation. All three solutions {-1, -1/3, 1} satisfy the equation.</p><p>∴ Answer: A</p>
Correct Answer: A