ache o valor de

por victor engler Dom 26 Ago 2012, 18:47

A = (x³ - y³ / x - y) - (x³ + y ³/x + y )

Sendo x>y>0 e que logx(base9) + log y (base9) = 1/2. Calcule o valor de logA(base√6).
gabarito= 2

por danjr5 Dom 26 Ago 2012, 19:17

CONDIÇÃO I:

$\\ A = \frac{x^3 - y^3}{x - y} - \frac{x^3 + y^3}{x + y} \\\\\\ A = \frac{(x - y)(x^2 + xy + y^2)}{x - y} - \frac{(x + y)(x^2 - xy + y^2)}{x + y} \\\\\\ A = x^2 + xy + y^2 - x^2 + xy - y^2 \\\\ \boxed{A = 2xy}$

CONDIÇÃO II:

$\\ log_9 \, x + log_9 \, y = \frac{1}{2} \\\\\\ log_9 (x\cdot y) = \frac{1}{2} \\\\ 9^{\frac{1}{2}} = xy \\\\ \boxed{xy = 3}$

Devemos calcular $\\ log_{\sqrt{6}} \, A$

Segue que

$\\ log_{\sqrt{6}} \, A = k \\\\ log_{\sqrt{6}} \, 2xy = k \\\\ log_{\sqrt{6}} \, 2 \cdot 3 = k \\\\ log_{\sqrt{6}} \, 6 = k \\\\ \sqrt{6}^k = 6 \\\\ 6^{\frac{k}{2}} = 6 \\\\ \frac{k}{2} = 1 \\\\ \boxed{\boxed{k = 2}}$

Espero ter ajudado.

Daniel F.