Exercițiul 223

E.223. Să se demonstreze formula: a,b(cd)=abcdab990.\overline{a,b(cd)} = \dfrac{\overline{abcd}-\overline{ab}}{990}.

MM, 24.04.2024
Soluție:

Metoda 1: Notăm a,b(cd)=x×10\overline{a,b(cd)} = x \quad | \times10
ab,(cd)=10x×100\overline{ab,(cd)} = 10x \quad | \times100
abcd,(cd)=1000x10x\overline{abcd,(cd)} = 1000x \quad | -10x
abcd,(cd)ab,(cd)=990x\overline{abcd,(cd)} - \overline{ab,(cd)} = 990x
abcdab=990xx=abcdab990.\overline{abcd} - \overline{ab} = 990x \Rightarrow \boxed{x=\dfrac{\overline{abcd}-\overline{ab}}{990}}.

Metoda 2: Ne folosimm de formula demonstrată anterior pentru 0,(ab).\overline{0,(ab)}.
a,b(cd)=110ab,(cd)=110(ab+0,(cd)=110(ab+cd99)=\overline{a,b(cd)} = \dfrac{1}{10} \cdot \overline{ab,(cd)} = \dfrac{1}{10} \cdot (\overline{ab} + \overline{0,(cd)} = \dfrac{1}{10} \cdot \Big(\overline{ab} + \dfrac{\overline{cd}}{99}\Big) =

=110100ab+cdab99=abcdab990.= \dfrac{1}{10} \cdot \dfrac{100 \cdot \overline{ab} + \overline{cd} - \overline{ab}}{99} = \dfrac{\overline{abcd}-\overline{ab}}{990}.