Question

cand transformam o fractie periodica in una ordinara de ce la numitor adaugam atati de 9 cate cifre avem in perioada ? primesc 10 daca o sa stiu luni​ k(ma refer de ce 9 si nu alta cifra)

Answer 1

Răspuns:

Explicație pas cu pas:

Pentru fractii periodice, simple

0,(3)

La numarator scriu toate cifrele, fara virgula, si se scad cele aflate inaintea virgulei.

La numitor scriu atatia de 9 cate cifre am in perioada.

Adica 0,(3)= 3/9

1,(33)= 133-1/99=132/99

14,(55)= (1455-14)/99= 1441/99

Pentru fractii periodice mixte

La numarator scriu toate cifrele, fara virgula, si scad cifrele aflate inaintea perioadei.

La numitor scriu atatia de 9 cate cifre sunt in perioada, apoi atatia de zero cate cifre sunt inaintea perioadei.

Exemplu:

1,2(3)= (12-1)/90=11/90

1,34(32)= (134324-134)9900=134190/9900

Acum incearca tu

3,(8)=

33,(45)=

0,12(4)=

11,22(77)=

Answer 2

$0,(3) = 0,33333\underset{de\,\,n\,\, ori}{\underbrace{...}}3,\quad n \to \infty\\ \\\\ =\dfrac{33333\underset{de\,\,n\,\, ori}{\underbrace{...}}3}{10000\underset{0\,\,de\,\,n\,\, ori}{\underbrace{...}}0} = \dfrac{3\cdot 11111\underset{de\,\,n\,\, ori}{\underbrace{...}}1}{10^n} =\\ \\\\ =\dfrac{3\cdot 9\cdot 11111\underset{de\,\,n\,\, ori}{\underbrace{...}}1}{9\cdot 10^n} = \dfrac{3\cdot 99999\underset{de\,\,n\,\, ori}{\underbrace{...}}9}{9\cdot 10^n} =$

$= \dfrac{3\cdot (10000\underset{0\,\,de\,\,n\,\, ori}{\underbrace{...}}0-1)}{9\cdot 10^n} = \dfrac{3\cdot (10^n-1)}{9\cdot 10^n} =\\ \\ =\dfrac{3\cdot 10^n -3}{9\cdot 10^n} = \dfrac{3\cdot 10^n}{9\cdot 10^n} - \dfrac{3}{9\cdot 10^n} = \dfrac{3}{9}- \dfrac{3}{9\cdot 10^{\infty}} = \\ \\ = \dfrac{3}{9}- \dfrac{3}{\infty} = \dfrac{3}{9}- 0 = \boxed{\dfrac{3}{9}} = \dfrac{1}{3}$

$\underline{\text{Pentru a generaliza putem spune ca, deoarece:}}\\\\121212 = 12\cdot 10101\\ 123123123 = 123\cdot 1001001\\ \\10101\cdot 99 = 999999\\ 1010101\cdot 99 = 99999999\\1001\cdot 999 = 999999\\\\ \\\underline{\text{Analog:}}\\ \\ \overline{a_1a_2a_3...a_ka_1a_2a_3...a_k\underset{de\,\, n\,\,ori}{\underbrace{.......}}a_1a_2a_3...a_k}=\\ \\= \overline{a_1a_2a_3...a_k} \cdot \underset{0\,\,de \,\,k-1\,\, ori,\,\,\,1\,\,de\,\,n\,\,ori}{\underbrace{10...010...010...01}}$

$\underline{\text{Demonstratie:}}\\ \\ 0,(\overline{a_1a_2a_3...a_k}) =\\ \\ = 0,\overline{a_1a_2a_3...a_ka_1a_2a_3...a_k\underset{de\,\,n\,\,ori}{\underbrace{.......}}a_1a_2a_3...a_k},\quad n\to \infty\\ \\\\ = \dfrac{\overline{a_1a_2a_3...a_ka_1a_2a_3...a_k\underset{de\,\,n\,\,ori}{\underbrace{.......}}a_1a_2a_3...a_k}}{10^{n\cdot k}} \\ \\\\ =\dfrac{\overline{a_1a_2a_3...a_k}\cdot\underset{0\,\,de\,\,k-1\,\,ori,\,\,\,1\,\,de\,\,n\,\,ori}{\underbrace{10...010...010...01}}}{10^{n\cdot k}}$

$=\dfrac{\overline{a_1a_2a_3...a_k}\cdot 999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot \underset{0\,\,de\,\,k-1\,\,ori,\,\,\,1\,\,de\,\,n\,\,ori}{\underbrace{10...010...010...01}}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{n\cdot k}}\\ \\ \\ = \dfrac{\overline{a_1a_2a_3...a_k}\cdot 999\underset{de\,\,n\cdot k\,\, ori}{\underbrace{...}9}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{n\cdot k}}$

$=\dfrac{\overline{a_1a_2a_3...a_k}\cdot (999\underset{de\,\,n \cdot k\,\, ori}{\underbrace{...}9}+1-1)}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{n\cdot k}}\\ \\\\ =\dfrac{\overline{a_1a_2a_3...a_k}\cdot (10^{n\cdot k}-1) }{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{n\cdot k}}$

$=\dfrac{\overline{a_1a_2a_3...a_k}\cdot 10^{n\cdot k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{n\cdot k}} - \dfrac{\overline{a_1a_2a_3...a_k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{n\cdot k}}$

$=\dfrac{\overline{a_1a_2a_3...a_k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}}- \dfrac{\overline{a_1a_2a_3...a_k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}\cdot 10^{\infty\cdot k}}\\ \\ \\ =\dfrac{\overline{a_1a_2a_3...a_k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}} - \dfrac{\overline{a_1a_2a_3...a_k}}{\infty}$

$=\dfrac{\overline{a_1a_2a_3...a_k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}} - 0 =\boxed{\dfrac{\overline{a_1a_2a_3...a_k}}{999\underset{de\,\,k\,\, ori}{\underbrace{...}9}}}$