Мисоли зеринро Нурмаҳмад Аминов пешниҳод кардааст.

Айнияти ададиро исбот мекунем:

\(\sqrt{\underbrace{11...1}_{2n+1\text{-то}} - \underbrace{11...1}_{2n\text{-то}}} = 1\underbrace{00...0}_{n\text{-то}}, \quad n \in N\)

Ҳал. Тарафи чапи айниятро табдил дода ҳосил мекунем:

\(\sqrt{\frac{1}{9}\cdot\underbrace{99...9}_{2n+1\text{-то}} - \frac{1}{9}\cdot\underbrace{99...9}_{2n\text{-то}}} = 1\underbrace{00...0}_{n\text{-то}}, \quad n \in N\)

Акнун аз баробарии

\((*) \quad 10^n - 1 = \underbrace{99...9}_{n\text{-то}}\)

истифода бурда, бори дигар табдил дода, ҳосил мекунем \((n \in N)\):

\(\sqrt{\frac{1}{9}\cdot \left(10^{2n+1} -1\right) - \frac{1}{9}\cdot\left(10^{2n} -1\right)} = 1\underbrace{00...0}_{n\text{-то}}\);

\(\sqrt{\frac{1}{9}\cdot 10^{2n+1} - \frac{1}{9} - \frac{1}{9}\cdot 10^{2n} + \frac{1}{9}} = 1\underbrace{00...0}_{n\text{-то}}\);

\(\sqrt{\frac{1}{9}\cdot 10^{2n+1} - \frac{1}{9}\cdot 10^{2n}} = 1\underbrace{00...0}_{n\text{-то}}\);

\(\sqrt{\frac{1}{9}\cdot 10^{2n}(10-1)} = 1\underbrace{00...0}_{n\text{-то}}\);

\(\sqrt{\frac{1}{9} \cdot 10^{2n} \cdot 9} = 1\underbrace{00...0}_{n\text{-то}}\);

\(\sqrt{10^{2n}} = 1\underbrace{00...0}_{n\text{-то}}\);

\(10^n = 1\underbrace{00...0}_{n\text{-то}}\);

\(1\underbrace{00...0}_{n\text{-то}} = 1\underbrace{00...0}_{n\text{-то}}\);