Question

Cum se desface paranteza (4a+3)×(b-1)
Sau mai bine urmatoarea problema pls:
Determinati nr nat nenule a si b, stiind ca 5b/4a+3=b-1/a
EXPLICATI PLS

Answer 1

"Determinati nr nat nenule a si b, stiind ca 5b/4a+3=b-1/a"

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Determinați numerele naturale nenule a și b,  știind că

5b/(4a+3) = (b-1)/a.

$\it \dfrac{5b}{4a+3} = \dfrac{b-1}{a} \Longrightarrow 5b\cdot a=(4a+3)(b-1) \Longrightarrow 5ab=4ab-4a+3b-3 \Longrightarrow\\ \\ \\ \Longrightarrow5ab-4ab+4a=3b-3 \Longrightarrow ab+4a=3b-b \Longrightarrow a(b+4)=3b-b\Longrightarrow\\ \\ \\ \Longrightarrow a=\dfrac{3b-3}{b+4}\ \ \ \ (1)$

$\it a\in\mathbb{N}^* \Longrightarrow \dfrac{3b-3}{b+4}\in\mathbb{N}^*\ \ \ \ \ (2)\\ \\ \\ (2) \Longrightarrow b+4|3b-3\ \ \ \ \ (3)\\ \\ \\ Dar,\ b+4|b+4 \Longrightarrow b+4|(b+4)\cdot3 \Longrightarrow b+4|3b+12\ \ \ \ \ (4)\\ \\ \\ (3),\ (4) \Longrightarrow b+4|3b+12-(3b-3) \Longrightarrow b+4|3b+12-3b+3 \Longrightarrow$

$\it \Longrightarrow b+4|15 \Longrightarrow b+4 \in \{1,\ 3,\ 5,\ 15\}\ \ \ \ \ (5)\\ \\ \\ b\in\mathbb{N}^* \Longrightarrow b>0|_{+4} \Longrightarrow b+4>4 \ \ \ \ \ (6)\\ \\ \\ (5),\ (6) \Longrightarrow b+4\in\{5,\ 15\}|_{-4} \Longrightarrow b\in\{1,\ 11\}$

$\it b=1 \stackrel{(1)}{\Longrightarrow} a=0\ \ (nu\ \ convine)\\ \\ b=11 \stackrel{(1)}{\Longrightarrow} a= 2$

Numerele cerute sunt:  a=2,  b=11.