Question

Salut, aveti o idee la problema 804 ???

Answer 1

Răspuns:

$\frac{x}{y} \in \Bigg\{1;4\Bigg\}$

Explicație pas cu pas:

$2lg(x-2y) = lg \:x + lg\:y\\\\lg\Big((x-2y)^2\Big) = lg \:xy\\\\ (x-2y)^2 = xy\\\\ x^2 -4xy + 4y^2 - xy = 0\\\\ x^2 - 5xy + 4y^2 = 0\\\\ k = \frac{x}{y}\iff x = ky\\\\ (ky)^2 - 5(ky)y + 4y^2 = 0\\ \\ k^2y^2 - 5ky^2 + 4y^2 = 0\\\\ \textrm{Rezolvam pentru k}\\\\ \Delta = {(5y^2)}^2 - 4\cdot y^2 \cdot 4y^2 = 25y^4 - 16y^4 = 9y^4\\\\ \sqrt{\Delta} = 3y^2\\\\k_{1,2} = \frac{5y^2 \pm 3y^2}{2y^2} \\\\ k_1 = \frac{2y^2}{2y^2} = 1\\\\ k_2 = \frac{8y^2}{2y^2} = 4\\\\ k \in \Bigg\{1;4\Bigg\} \implies \frac{x}{y} \in \Bigg\{1;4\Bigg\}$

Answer 2

$\it 2lg(x-2y)=lgx+lgy \Rightarrow lg(x-2y)^2=lgxy \Rightarrow (x-2y)^2=xy \Rightarrow\\ \\ \\ \Rightarrow x^2-4xy+4y^2=xy|_{-xy} \Rightarrow x^2-5xy+4y^2=0|_{:y^2} \Rightarrow \Big(\dfrac{x}{y}\Big)^2-5\dfrac{x}{y}+4=0\Rightarrow \\ \\ \\ \Big(\dfrac{x}{y}\Big)^2-\dfrac{x}{y}-4\dfrac{x}{y}+4=0\Rightarrow \dfrac{x}{y}\Big(\dfrac{x}{y}-1\Big)-4\Big(\dfrac{x}{y}-1\Big)=0\Rightarrow \Big(\dfrac{x}{y}-1\Big)\Big(\dfrac{x}{y}-4\Big)=0\Rightarrow$

$\it \Rightarrow \begin{cases}\it \dfrac{x}{y}-1=0 \Rightarrow \dfrac{x}{y}=1 \\ \\ sau\\ \\ \it \dfrac{x}{y}-4=0 \Rightarrow \dfrac{x}{y}=4\end{cases}\ \ \ \Rightarrow\ \dfrac{x}{y} \in \{1,\ 4\}$