How to solve this nice math Exponential algebra problem | Olympiad Question | x=?

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[@HuzaifaKhan-m3i]

❤❤❤❤

[@gourivulchi7700]

Nice one

[@AfsarSaid-f8s]

Good question

[@walterwen2975]

Olympiad Question: √(1 - 1/x) + √(x - 1/x) = x; x =? x ≠ 0; [√(x - 1/x)]² = [x - √(1 - 1/x)]², x - 1/x = x² - 2x√(1 - 1/x) + 1 - 1/x 2x√(1 - 1/x) = x² - x + 1, 2√(x² - x) = x² - x + 1, Let: y = x² - x; 2√y = y + 1 (2√y)² = (y + 1)², 4y = y² + 2y + 1, y² - 2y + 1 = 0, (y - 1)² = 0, y = 1; Double root y = x² - x = 1, x² - x - 1 = 0, x = (1 ± √5)/2 Answer check: x = (1 ± √5)/2: x² - x - 1 = 0, x² - 1 = x [√(1 - 1/x) + √(x - 1/x)]² = (1 - 1/x) + 2√[(1 - 1/x)(x - 1/x)] + (x - 1/x) = 1 + x - 2/x + 2√(x - 1 - 1/x + 1/x²) = (x² + x - 2)/x + 2√[(x² - x - 1)/x + 1/x²] = (x + x - 1)/x + 2√[(x - x)/x + 1/x²)] = 2 - 1/x + 2/x = 2 + 1/x = 2 + 2/(1 ± √5) = 2 + [2(- 1 ± √5)]/4 = 2 + (- 1 ± √5)/2 = (3 ± √5)/2 = (6 ± 2√5)/4 = [1 ± 2√5 + (√5)²]/2² = [(1 ± √5)/2]² = x²; Confirmed Final answer: x = (1 + √5)/2 or x = (1 -√5)/2; Double roots

[@RealQinnMalloryu4]

(1)^2 ➖ (1)^2/(x)^2+(x)^2 ➖ (1)^2/(x)^2={1 ➖ 1}/x^2+{x^2 ➖ 1}/x^2={0+0 ➖ }/x^2+{x^0+x^0 ➖ x^0+x^0 ➖ }/x^2={1/x^2+x^2/x^2}=1x^2/x^4=1x^2 (x ➖ 2x+1).