Question

determinati perechile de numere naturale (a,b), stiind ca a-1=10/b+3

Answer 1

a-1=10/(b+3)
(a-1)(b+3)=10
Scriem toate solutiile posibile de al descompune pe 10 in produs de doi factori:
10=10*1
10=1*10
10=5*2
10=2*5
Si luam fiecare caz pe rand:
{ a-1=10 ⇒{ a=11
{ b+3=1  ⇒{ b=-2∉N deci solutia nu e buna

{a-1=1    ⇒{a=2
{b+3=10 ⇒{b=7

{a-1=2  ⇒{a=3
{b+3=5 ⇒{b=2

{a-1=5  ⇒{a=6
{b+3=2 ⇒{b=-1 ∉N deci nici asta nu convine
S:(a,b)∈{(2,7) , (3,2)}

Answer 2

[tex]a-1 = \dfrac{10}{b}+3\\ \\ \\ a \in \mathbb_N $ $ \Rightarrow a-1 \in \mathbb_N $ $ \Rightarrow \\ \\ \Rightarrow \dfrac{10}{b} +3 \in \mathbb_N $ $ \Rightarrow \\ \\ \Rightarrow b | 10 \Rightarrow b \in D_{10} \Rightarrow b \in \Big\{1,2,5,10\Big\}\Big|^{-1} \Rightarrow \\ \\\Rightarrow \dfrac{1}{b} \in \Big\{1,\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{10}\Big\} \Big|\cdot 10 \Rightarrow \\ \\ \Rightarrow \dfrac{10}{b} \in \Big\{10,5,2,1\Big\}\Big|+3 \Rightarrow \\ \\ \Rightarrow \dfrac{10}{b}+3 \in \Big\{13,8,5,4\Big\} \Rightarrow \\ \\ \Rightarrow a-1 \in \Big\{13,8,5,4\Big\} |+1 \Rightarrow \\ \\ \Rightarrow \boxed{a\in \Big\{14,9,6,5\Big\}} \\ ~~~~~\boxed{b \in \Big\{1,2,5,10\Big\}}\\ \\ \\ \Rightarrow (a,b) \in \Big\{(14,1);(9,2);(6,5);(5,10)\Big\} [/tex]