If a-b-0 are two real numbers- the value of -a b-a-b -a b-a-b -a b-a-b -a b- is

√(ab + (a b)√(ab + (a b) √(ab + (a b) √(ab +...)))) = t √(ab + (a b)t) = t ab + (a b)t = t^2 t^2 (a b) t ab = 0 t^2 at + bt ab = 0 1/2 (t a) + b(t a) ...

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If a>b>0 are ...

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If a > b > 0 are two real numbers, the value of $\sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + . . .}}}}$ is

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$\sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + . . .}}}} = t \sqrt{a b + (a - b) t} = t a b + (a - b) t = t^{2} t^{2} - (a - b) t - a b = 0 t^{2} - a t + b t - a b = 0 \frac{1}{2} (t - a) + b (t - a) = 0 (t + b) (t - a) = 0 t = - b, a$

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If a > b > 0 are two real numbers, the value of √ab+(a−b)√ab+(a−b)√ab+(a−b)√ab+... is

√ab+(a−b)√ab+(a−b)√ab+(a−b)√ab+...=t√ab+(a−b)t=tab+(a−b)t=t2t2−(a−b)t−ab=0t2−at+bt−ab=012(t−a)+b(t−a)=0(t+b)(t−a)=0t=−b,a

If a > b > 0 are two real numbers, the value of $\sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + . . .}}}}$ is

$\sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + (a - b) \sqrt{a b + . . .}}}} = t \sqrt{a b + (a - b) t} = t a b + (a - b) t = t^{2} t^{2} - (a - b) t - a b = 0 t^{2} - a t + b t - a b = 0 \frac{1}{2} (t - a) + b (t - a) = 0 (t + b) (t - a) = 0 t = - b, a$