19. Sic \( \frac{1}{x}+\frac{1}{y}=\frac{4}{x+y} \) Calcallar: \( E=\frac{x^{2}-y^{2}}{x y}+\frac{x y}{x+y}+\frac{(x+y)^{2}}{x^{2}} \)
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To solve the equation \( \frac{1}{x}+\frac{1}{y}=\frac{4}{x+y} \), we can first manipulate the given equation. Rewrite the left side to have a common denominator: \[ \frac{y + x}{xy} = \frac{4}{x+y} \] Cross-multiplying gives: \[ (x+y)(x+y) = 4xy \] This simplifies to: \[ x^2 + 2xy + y^2 = 4xy \quad \Rightarrow \quad x^2 - 2xy + y^2 = 0 \quad \Rightarrow \quad (x - y)^2 = 0 \] Thus, we find \( x = y \). Now substituting \( x = y \) into \( E \): \[ E = \frac{x^2 - x^2}{x^2} + \frac{x^2}{2x} + \frac{(2x)^2}{x^2} \] \[ E = 0 + \frac{x}{2} + 4 = \frac{x}{2} + 4 \] We can evaluate \( E \) with any specific value for \( x \) (or \( y\)) since they are equal, leading to a cleaner solution based on substitution!