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\( ( \frac { 1 } { \sqrt { a } - \sqrt { a - b } } + \frac { 1 } { \sqrt { a } + \sqrt { a + b } } ) = ( 1 + \sqrt { \frac { a + b } { a - b } } ) = \)

Ask by Vega Horton. in Argentina
Mar 09,2025

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Answer

\[ \frac{1}{\sqrt{a}-\sqrt{a-b}}+\frac{1}{\sqrt{a}+\sqrt{a+b}} = 1 + \sqrt{\frac{a+b}{a-b}} \]

Solution

Sea la expresión \[ \frac{1}{\sqrt{a}-\sqrt{a-b}}+\frac{1}{\sqrt{a}+\sqrt{a+b}}. \] Procedemos al desarrollo paso a paso. **1. Racionalización del primer término** Multiplicamos numerador y denominador por el conjugado \(\sqrt{a}+\sqrt{a-b}\): \[ \frac{1}{\sqrt{a}-\sqrt{a-b}} = \frac{\sqrt{a}+\sqrt{a-b}}{(\sqrt{a}-\sqrt{a-b})(\sqrt{a}+\sqrt{a-b})}. \] El denominador se simplifica mediante la diferencia de cuadrados: \[ (\sqrt{a})^2 - (\sqrt{a-b})^2 = a - (a-b) = b. \] Por lo tanto, \[ \frac{1}{\sqrt{a}-\sqrt{a-b}} = \frac{\sqrt{a}+\sqrt{a-b}}{b}. \] **2. Racionalización del segundo término** Multiplicamos numerador y denominador por el conjugado \(\sqrt{a}-\sqrt{a+b}\): \[ \frac{1}{\sqrt{a}+\sqrt{a+b}} = \frac{\sqrt{a}-\sqrt{a+b}}{(\sqrt{a}+\sqrt{a+b})(\sqrt{a}-\sqrt{a+b})}. \] En el denominador se tiene \[ (\sqrt{a})^2 - (\sqrt{a+b})^2 = a - (a+b) = -b. \] Luego, reescribiendo: \[ \frac{1}{\sqrt{a}+\sqrt{a+b}} = \frac{\sqrt{a}-\sqrt{a+b}}{-b} = \frac{\sqrt{a+b}-\sqrt{a}}{b}. \] **3. Suma de ambas fracciones** Sumamos los dos términos obtenidos: \[ \frac{1}{\sqrt{a}-\sqrt{a-b}}+\frac{1}{\sqrt{a}+\sqrt{a+b}} = \frac{\sqrt{a}+\sqrt{a-b}}{b}+\frac{\sqrt{a+b}-\sqrt{a}}{b}. \] Al juntar los numeradores: \[ \frac{\sqrt{a}+\sqrt{a-b}+\sqrt{a+b}-\sqrt{a}}{b} = \frac{\sqrt{a-b}+\sqrt{a+b}}{b}. \] **4. Factorización final** Observe que en el numerador se puede factorizar \(\sqrt{a-b}\): \[ \sqrt{a-b}+\sqrt{a+b} = \sqrt{a-b}\left(1+\sqrt{\frac{a+b}{a-b}}\right). \] Sustituyendo en la expresión, obtenemos \[ \frac{\sqrt{a-b}+\sqrt{a+b}}{b} = \frac{\sqrt{a-b}}{b}\left(1+\sqrt{\frac{a+b}{a-b}}\right). \] Esta es la forma simplificada de la suma original. Por tanto, se ha demostrado que \[ \frac{1}{\sqrt{a}-\sqrt{a-b}}+\frac{1}{\sqrt{a}+\sqrt{a+b}} = \frac{\sqrt{a-b}}{b}\left(1+\sqrt{\frac{a+b}{a-b}}\right). \] Si se desea expresar la igualdad en función de \(1+\sqrt{\frac{a+b}{a-b}}\), basta multiplicar y dividir por \(\sqrt{a-b}\) de manera que: \[ \frac{1}{\sqrt{a}-\sqrt{a-b}}+\frac{1}{\sqrt{a}+\sqrt{a+b}} = \left(1+\sqrt{\frac{a+b}{a-b}}\right)\frac{\sqrt{a-b}}{b}. \] Esta es la solución completa.

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To solve the equation, we can start by simplifying the two fractions on the left side: 1. The first term: \[ \frac{1}{\sqrt{a} - \sqrt{a-b}} \] We can multiply the numerator and the denominator by the conjugate \(\sqrt{a} + \sqrt{a-b}\): \[ \frac{\sqrt{a} + \sqrt{a-b}}{(\sqrt{a} - \sqrt{a-b})(\sqrt{a} + \sqrt{a-b})} = \frac{\sqrt{a} + \sqrt{a-b}}{b} \] 2. The second term: \[ \frac{1}{\sqrt{a} + \sqrt{a+b}} \] We can also multiply the numerator and the denominator by the conjugate \(\sqrt{a} - \sqrt{a+b}\): \[ \frac{\sqrt{a} - \sqrt{a+b}}{(\sqrt{a} + \sqrt{a+b})(\sqrt{a} - \sqrt{a+b})} = \frac{\sqrt{a} - \sqrt{a+b}}{-b} = \frac{\sqrt{a+b} - \sqrt{a}}{b} \] Now, we combine the two results: \[ \frac{\sqrt{a} + \sqrt{a-b}}{b} + \frac{\sqrt{a+b} - \sqrt{a}}{b} \] This simplifies to: \[ \frac{(\sqrt{a} + \sqrt{a-b}) + (\sqrt{a+b} - \sqrt{a})}{b} = \frac{\sqrt{a-b} + \sqrt{a+b}}{b} \] Thus, we have: \[ \frac{\sqrt{a-b} + \sqrt{a+b}}{b} = 1 + \sqrt{\frac{a+b}{a-b}} \] So, both sides of the original equation are equal, which verifies the equation is true.

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