Question

Determinati numarul natural nenu n stiind ca 2x=3y=6z=(x+z+y)×n . Varog ajutatima

Answer 1

[tex]2x=3y=6z=(x+y+z)\cdot n\\ \text{Daca facem un artificiu de calcul obtinem:}\\ \dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{6}}=\dfrac{x+y+z}{\frac{1}{n}} (1)\\ \text{Dar din egalitatea proportiilor avem ca:} \dfrac{x}{\frac{1}{2}}=\dfrac{y}{\frac{1}{3}}=\dfrac{z}{\frac{1}{6}}=\dfrac{x+y+z}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}} (2)\\ \text{Din (1) si (2) obtinem:}\\ \\ \dfrac{1}{n}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\\ \\ \dfrac{1}{n}=\dfrac{3+2+1}{6}\\ [/tex]
$\dfrac{1}{n}=\dfrac{6}{6}=1\Rightarrow \boxed{n=1}$