exercitiu cls a 4-a

oana5 · Mesaj de **oana5** » 28 Aug 2015, 17:33

1+21:{1+[a+(a:4-2)2]:7}=8

adrianS · Mesaj de **adrianS** » 29 Aug 2015, 14:11

Buna ziua
Scriem:
$Exp=1+\dfrac{21}{1+[a+(\dfrac{a}{4}-2)2]\cdot\dfrac{1}{7}\\ =1+\dfrac{147}{7+[a+(\dfrac{a}{4}-2)2]\\ =1+\dfrac{147}{7+[a+\dfrac{a-8}{2}]\\ =1+\dfrac{147}{7+\dfrac{2a+a-8}{2}\\ =1+\dfrac{294}{3a+6}=8$
De aici :
$3a+6+294=24a+48\\ 21a=294+6-48=252\\ a=\dfrac{252}{21}=12$

oana5 · Mesaj de **oana5** » 29 Aug 2015, 20:39

Wow. Multumesc mult. Nu reuseam sa-i dau de cap. Mă blocam mereu pe la mijlocul rezolvării.

Mesaj de **ex-admin** » 30 Aug 2015, 00:14

La clasa a IV-a e mai potrivita urmatoarea abordare:

Pas cu pas, identificam operatia principala (ultima care s-ar efectua daca toate numerele din expresia de calculat ar fi cunoscute) si aflam operandul necunoscut:

$\begin{array}{l} \,\,\,\,\,1 + \underbrace {21:\{ 1 + [a + (a:4 - 2) \cdot 2]:7\} }_? = 8\, \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + ? = 8 \Rightarrow ? = 7 \\ \Rightarrow \\ \,\,\,\,\,\,21:\underbrace {\{ 1 + [a + (a:4 - 2) \cdot 2]:7\} }_? = 7 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21:? = 7 \Rightarrow ? = 3 \\ \Rightarrow \\ \,\,\,\,\,\,1 + \underbrace {[a + (a:4 - 2) \cdot 2]:7}_? = 3 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + ? = 3 \Rightarrow ? = 2 \\ \Rightarrow \\ \,\,\,\,\,\,\,\underbrace {[a + (a:4 - 2) \cdot 2]}_?:7 = 2 \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,?:7 = 2 \Rightarrow ? = 14 \\ \Rightarrow \\ \,\,\,\,\,\,a + (a:4 - 2) \cdot 2 = 14 \\ \end{array}$

Acum operatia principala este cea de adunare dar ambii operanzi (termeni, in cazul de fata) sunt necunoscuti, astfel ca nu mai putem continua dupa procedeul descris anterior.

Daca pana acum am avut de a face cu un tip des intalnit de exercitiu, abia acum, aflarea lui $a$ din $a + (a:4 - 2) \cdot 2 = 14$ ne cere putina perspicacitate!

Vom nota sfertul lui $a$ cu $b$ . Atunci $a = 4 \cdot b$ si ecuatia devine $4 \cdot b + (b - 2) \cdot 2 = 14$

Finalizarea:

$\begin{array}{l} 4 \cdot b + (b - 2) \cdot 2 = 14 \Rightarrow \\ 4b + 2b - 4 = 14 \Rightarrow \\ 6b - 4 = 14 \Rightarrow \\ 6b = 18 \Rightarrow \\ b = 3 \Rightarrow \\ a = 12 \\ \end{array}$

adrianS · Mesaj de **adrianS** » 30 Aug 2015, 09:55

Buna ziua
Asa este!
Rezolvarea oferita de dumneavoastra pare a fi mai potrivita pentru nivelul clasei a sasea!
Interesant mi se pare procedeul de reducere al cifrelor ex.8-1=7,3-1=2 etc.