Question

Determinati nr întregi a si b, astfel incat:
/=supra
a) 1/a+b=b-3/5

b) a+1/2=3/b-2

Va rog ! Am nevoie de ajutor!

Answer 1

$\it \dfrac{1}{a+b}=\dfrac{b-3}{5} \Rightarrow a+b =\dfrac{5}{b-3} \Rightarrow a=\dfrac{5}{b-3}-b \ \ \ \ \ \ (1)\\ \\ \\ a\in\mathbb{Z}, \ \ b\in\mathbb{Z} \stackrel{(1)}{\Longrightarrow} \dfrac{5}{b-3} \in\mathbb{Z} \Rightarrow b-3 \in\{-5,\ -1,\ 1,\ 5\}|_{+3} \Rightarrow \\ \\ \\ \Rightarrow b\in\{-2,\ 2,\ 4,\ 8\}\ \ \ \ \ \ (2)$

$\it (1), (2) \Rightarrow \begin{cases}\it a=\dfrac{5}{-2-3} -(-2) \Rightarrow a=1 \\ \\ \\ \it a=\dfrac{5}{2-3} -2 \Rightarrow a=-7\\ \\ \\ \it a=\dfrac{5}{4-3} -4\Rightarrow a=1 \\ \\ \\ \it a=\dfrac{5}{8-3} -8 \Rightarrow a=-7 \end{cases}\\ \\ \\ (a,\ b) \in\{(1,\ -2),\ (-7,\ 2),\ (1,\ 4),\ (-7,\ 8)\}$

b)

$\it \dfrac{a+1}{2} =\dfrac{3}{b-2} \Rightarrow (a+1)(b-2) =6 =-6\cdot(-1)=-1\cdot(-6)=1\cdot6=6\cdot1=\\ \\ \\ =-2\cdot(-3) =-3\cdot(-2) =2\cdot3=3\cdot2\\ \\ \\ a+1\in\{-6,\ -1,\ 1,\ 6,\ -2,\ -3,\ 2,\ 3\}|_{-1} \Rightarrow a\in\{-7,\ -2,\ 0,\ 5,\ -3,\ -4,\ 1,\ 2\}$

$\it b-2\in\{-1,\ -6,\ 6,\ 1,\ -3,\ -2,\ 3,\ 2\}|_{+2} \Rightarrow\\ \\ \\ \Rightarrow b\in\{1,\ -4,\ 8,\ 3,\ -1,\ 0,\ 5,\ 4\}$

$\it (a,\ b) \in\{(-7,\ 1), (-2,\ -4), (0,\ 8), (5,\ 3), (-3,\ -1),( -4,\ 0),(1,5),(2,4)\}$