x-1-x-1-1-x-1-1-x-

Question Number 202638 by depressiveshrek last updated on 31/Dec/23

\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} = 1 - \frac{1}{x}

Commented by AST last updated on 31/Dec/23

x = 1 . S u p p o s e x \neq 1, l e t a = \sqrt{x - \frac{1}{x}}; b = \sqrt{1 - \frac{1}{x}}

a - b = 1 - \frac{1}{x}; {(a + b)}^{2} = 1 + x

a^{2} + b^{2} = x + 1 - \frac{2}{x} \Rightarrow 2 a b = \frac{2}{x}

{(a + b)}^{2} - {(a - b)}^{2} = 4 a b \Rightarrow x - \frac{1}{x^{2}} + \frac{2}{x} = \frac{4}{x}

\Rightarrow x^{3} - 2 x - 1 = 0 \Rightarrow x = \frac{1 + \sqrt{5}}{2}

Commented by Rasheed.Sindhi last updated on 31/Dec/23

S i r m r W “ c a n {s e e}^{″} g e o m e t r y i n t h e q u e s t i o n .

Commented by AST last updated on 31/Dec/23

Commented by AST last updated on 31/Dec/23

O R a^{2} - b^{2} = (a - b) (a + b) = x - 1

\Rightarrow a - b = \frac{x - 1}{\sqrt{1 + x}} = \frac{x - 1}{x} \Rightarrow x = 1 o r x = \sqrt{x + 1}

\Rightarrow x^{2} - x - 1 = 0 \Rightarrow x = \frac{1 + \sqrt{5}}{2}

Answered by Rasheed.Sindhi last updated on 01/Jan/24

\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} = 1 - \frac{1}{x}

\sqrt{\frac{x^{2} - 1}{x}} - \sqrt{\frac{x - 1}{x}} - \frac{x - 1}{x} = 0

\sqrt{\frac{x - 1}{x} (x + 1)} - \sqrt{\frac{x - 1}{x}} - {(\sqrt{\frac{x - 1}{x}})}^{2} = 0

\sqrt{\frac{x - 1}{x}} (\sqrt{x + 1} - 1 - \sqrt{\frac{x - 1}{x}}) = 0

\sqrt{\frac{x - 1}{x}} = 0 ∣ \sqrt{x + 1} - 1 - \sqrt{\frac{x - 1}{x}} = 0

\frac{x - 1}{x} = 0 ∣ \sqrt{x + 1} - \sqrt{\frac{x - 1}{x}} = 1

x = 1 ✓ ∣ (x + 1) + (\frac{x - 1}{x}) - 2 \sqrt{\frac{x^{2} - 1}{x}} = 1

x + 1 + (1 - \frac{1}{x}) - 2 \sqrt{x - \frac{1}{x}} = 1

1 + x - \frac{1}{x} - 2 \sqrt{x - \frac{1}{x}} = 0

1 + {(\sqrt{x - \frac{1}{x}})}^{2} - 2 \sqrt{x - \frac{1}{x}} = 0

y^{2} - 2 y + 1 = 0

y = 1

\sqrt{x - \frac{1}{x}} = 1

x^{2} - x - 1 = 0

x = \frac{1 \pm \sqrt{1 + 4}}{2}

= \frac{1 \pm \sqrt{5}}{2}

Commented by AST last updated on 31/Dec/23

8 t h l i n e : x + 1 + (1 - \frac{1}{x}) - 2 \sqrt{x - \frac{1}{x}} = 0

Commented by Rasheed.Sindhi last updated on 31/Dec/23

T h a n k s s i r, I^{'} v e c o r r e c t e d .

Commented by Rasheed.Sindhi last updated on 02/Jan/24

R e C o r r e c t e d . N o w t h e a n s w e r i s

r i g h t!