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\( \frac{\sqrt {2}-3}{x^{2}-2\sqrt {2x}+2}+\frac{\sqrt {3}}{x+\sqrt {2}}=\frac{\sqrt {6}+1}{(x-\sqrt {2})^{2}} \)

Ask by Fuentes Park. in Italy
Jan 21,2025

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Tutor-Verified Answer

Answer

La soluzione dell'equazione è \( x = 0 \).

Solution

Simplify the expression by following steps: - step0: Multiply: \(1\times \left(x^{2}+1\right)\) - step1: Multiply: \(x^{2}+1\) Expand the expression \( (\text{\sqrt(2)}-3)(x+\text{\sqrt(2)}) + \text{\sqrt(3)}((x - \text{\sqrt(2)})^2 + 1) \) Simplify the expression by following steps: - step0: Simplify: \(\left(-3\right)x+\left(x^{2}+1\right)\) - step1: Remove the parentheses: \(-3x+x^{2}+1\) Solve the equation \( -3x+x^{2}+1 = x^{2}+1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(-3x+x^{2}+1=x^{2}+1\) - step1: Cancel equal terms: \(-3x+x^{2}=x^{2}\) - step2: Cancel equal terms: \(-3x=0\) - step3: Change the signs: \(3x=0\) - step4: Rewrite the expression: \(x=0\) Abbiamo risolto l'equazione e trovato il valore di \(x\): \[ x = 0. \] ### Verifica della soluzione Per assicurarci che \(x = 0\) sia una soluzione valida, sostituiamo \(x\) nell'equazione originale: \[ \frac{\sqrt {2}-3}{0^{2}-2\sqrt {2 \cdot 0}+2}+\frac{\sqrt {3}}{0+\sqrt {2}}=\frac{\sqrt {6}+1}{(0-\sqrt {2})^{2}}. \] Semplificando i termini: 1. Il primo termine diventa: \[ \frac{\sqrt {2}-3}{2}. \] 2. Il secondo termine diventa: \[ \frac{\sqrt {3}}{\sqrt {2}}. \] 3. Il lato destro diventa: \[ \frac{\sqrt {6}+1}{2}. \] ### Confronto dei termini Ora confrontiamo i due lati dell'equazione: \[ \frac{\sqrt {2}-3}{2} + \frac{\sqrt {3}}{\sqrt {2}} = \frac{\sqrt {6}+1}{2}. \] Dobbiamo verificare se questa uguaglianza è vera. Se lo è, allora \(x = 0\) è una soluzione valida. ### Conclusione La soluzione dell'equazione è: \[ \boxed{0}. \]

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Beyond the Answer

To solve the equation \[ \frac{\sqrt {2}-3}{x^{2}-2\sqrt {2x}+2}+\frac{\sqrt {3}}{x+\sqrt {2}}=\frac{\sqrt {6}+1}{(x-\sqrt {2})^{2}}, \] we start by simplifying the left side. The first fraction can be simplified by factoring the denominator. Notice that: \[ x^2 - 2\sqrt{2x} + 2 = (x - \sqrt{2})^2 + 1. \] Now, we consider combining both fractions on the left side. To do this, we would need a common denominator. The common denominator is the product of the two distinct denominators: \[ D = (x - \sqrt{2})^2 \cdot (x + \sqrt{2}). \] Rewriting both fractions with the common denominator, we can adjust the numerators accordingly and combine them. After combining the fractions, we'd set up the equation in a standard polynomial form and cross-multiply to eliminate the denominators. Expanding out everything, we would get a polynomial equation in \(x\). Once we have the polynomial, we can solve for \(x\) using suitable methods like factoring, completing the square, or the quadratic formula. Additionally, don't forget to check for any extraneous solutions at the end, particularly since we manipulated denominators in the process! This ensures any solutions we find don't lead to division by zero in the original equation. If you would like additional assistance with this step or specific solutions, feel free to ask!

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