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Home » 3録 MSUSIS Secrets Everyone’s Hiding—Don’t Miss This Transformative Breakthrough! - Blask

3録 MSUSIS Secrets Everyone’s Hiding—Don’t Miss This Transformative Breakthrough! - Blask

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Case 1: \( y > 1 \), both numerator and denominator positive.

📌 Case 2: \( -1 \leq y < 1 \), numerator non-negative, denominator negative → expression negative, not allowed.

📌 Case 3: \( y < -1 \), numerator and denominator both negative → positive.

📌 But if \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), valid. However, \( x^2 = \frac{y+1}{y-1} \geq 0 \) requires this fraction non-negative, which it is.

📌 But we must check if all such \( y \) are attainable. From \( x^2 = \frac{y+1}{y-1} \), for any \( y \ne 1 \), there's a real \( x \) if the right-hand side is non-negative — which holds when \( y > 1 \) or \( y \leq -1 \).

📌 But when \( y = -1 \), \( x^2 = 0 \Rightarrow x = 0 \), valid.

📌 As \( y \to 1^- \), \( \frac{y+1}{y-1} \to -\infty \), invalid.

📌 When \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), e.g., \( y = -2 \Rightarrow x^2 = \frac{-1}{-3} = \frac{1}{3} \), real \( x \).

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