▶️ Ejercicio 250 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXXI

R a d i c a l e s
División de radicales cuando el divisor es compuesto

Ejercicio 250

Dividir:

$\begin{matrix} \sqrt{2} e n t r e \sqrt{2} + \sqrt{3} \\ \frac{\sqrt{2}}{\sqrt{2} + \sqrt{3}} & = \frac{\sqrt{2}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} \\ = \frac{2 - \sqrt{6}}{2 - 3} \\ = \frac{2 - \sqrt{6}}{- 1} \\ = \sqrt{6} - 2 \end{matrix}$
$\begin{matrix} \sqrt{3} e n t r e \sqrt{3} - 2 \sqrt{5} \\ \frac{\sqrt{3}}{\sqrt{3} - 2 \sqrt{5}} & = \frac{\sqrt{3}}{\sqrt{3} - 2 \sqrt{5}} \times \frac{\sqrt{3} + 2 \sqrt{5}}{\sqrt{3} + 2 \sqrt{5}} \\ = \frac{3 + 2 \sqrt{15}}{{(\sqrt{3})}^{2} - {(2 \sqrt{5})}^{2}} \\ = \frac{3 + 2 \sqrt{15}}{3 - 4 (5)} \\ = \frac{3 + 2 \sqrt{15}}{3 - 20} \\ = - \frac{3 + 2 \sqrt{15}}{17} \end{matrix}$
$\begin{matrix} 2 + \sqrt{5} e n t r e 1 - \sqrt{5} \\ \frac{2 + \sqrt{5}}{1 - \sqrt{5}} & = \frac{2 + \sqrt{5}}{1 - \sqrt{5}} \times \frac{1 + \sqrt{5}}{1 + \sqrt{5}} \\ = \frac{2 + \sqrt{5} + 2 \sqrt{5} + 5}{1 - {(\sqrt{5})}^{2}} \\ = \frac{7 + 3 \sqrt{5}}{1 - 5} \\ = \frac{7 + 3 \sqrt{5}}{- 4} \\ = - \frac{7 + 3 \sqrt{5}}{4} \end{matrix}$
$\begin{matrix} \sqrt{2} + \sqrt{5} e n t r e \sqrt{2} - \sqrt{5} \\ \frac{\sqrt{2} + \sqrt{5}}{\sqrt{2} - \sqrt{5}} & = \frac{\sqrt{2} + \sqrt{5}}{\sqrt{2} - \sqrt{5}} \times \frac{\sqrt{2} + \sqrt{5}}{\sqrt{2} + \sqrt{5}} \\ = \frac{{(\sqrt{2} + \sqrt{5})}^{2}}{{(\sqrt{2})}^{2} - {(\sqrt{5})}^{2}} \\ = \frac{2 + 2 \sqrt{10} + 5}{2 - 5} \\ = - \frac{7 + 2 \sqrt{10}}{3} \end{matrix}$
$\begin{matrix} 2 \sqrt{3} - \sqrt{7} e n t r e \sqrt{3} + \sqrt{7} \\ \frac{2 \sqrt{3} - \sqrt{7}}{\sqrt{3} + \sqrt{7}} & = \frac{2 \sqrt{3} - \sqrt{7}}{\sqrt{3} + \sqrt{7}} \times \frac{\sqrt{3} - \sqrt{7}}{\sqrt{3} - \sqrt{7}} \\ = \frac{2 \sqrt{3^{2}} - \sqrt{21} - 2 \sqrt{21} + \sqrt{7^{2}}}{{(\sqrt{3})}^{2} - {(\sqrt{7})}^{2}} \\ = \frac{2 (3) - 3 \sqrt{21} + 7}{3 - 7} \\ = \frac{6 - 3 \sqrt{21} + 7}{- 4} \\ = \frac{3 \sqrt{21} - 13}{4} \end{matrix}$
$\begin{matrix} \sqrt{6} + 2 \sqrt{5} e n t r e 2 \sqrt{6} - \sqrt{5} \\ \frac{\sqrt{6} + 2 \sqrt{5}}{2 \sqrt{6} - \sqrt{5}} & = \frac{\sqrt{6} + 2 \sqrt{5}}{2 \sqrt{6} - \sqrt{5}} \times \frac{2 \sqrt{6} + \sqrt{5}}{2 \sqrt{6} + \sqrt{5}} \\ = \frac{2 (6) + 4 \sqrt{30} + \sqrt{30} + 2 (5)}{{(2 \sqrt{6})}^{2} - {(\sqrt{5})}^{2}} \\ = \frac{12 + 5 \sqrt{30} + 10}{4 (6) - 5} \\ = \frac{22 + 5 \sqrt{30}}{19} \end{matrix}$
$\begin{matrix} 5 \sqrt{2} + 3 \sqrt{3} e n t r e 3 \sqrt{2} - 4 \sqrt{3} \\ \frac{5 \sqrt{2} + 3 \sqrt{3}}{3 \sqrt{2} - 4 \sqrt{3}} & = \frac{5 \sqrt{2} + 3 \sqrt{3}}{3 \sqrt{2} - 4 \sqrt{3}} \times \frac{3 \sqrt{2} + 4 \sqrt{3}}{3 \sqrt{2} + 4 \sqrt{3}} \\ = \frac{15 \sqrt{2^{2}} + 9 \sqrt{6} + 20 \sqrt{6} + 12 \sqrt{3^{2}}}{{(3 \sqrt{2})}^{2} - {(4 \sqrt{3})}^{2}} \\ = \frac{15 (2) + 29 \sqrt{6} + 12 (3)}{9 (2) - 16 (3)} \\ = \frac{30 + 29 \sqrt{6} + 36}{18 - 48} \\ = - \frac{66 + 29 \sqrt{6}}{30} \end{matrix}$
$\begin{matrix} \sqrt{7} - 2 \sqrt{11} e n t r e 2 \sqrt{7} + \sqrt{11} \\ \frac{\sqrt{7} - 2 \sqrt{11}}{2 \sqrt{7} + \sqrt{11}} & = \frac{\sqrt{7} - 2 \sqrt{11}}{2 \sqrt{7} + \sqrt{11}} \times \frac{2 \sqrt{7} - \sqrt{11}}{2 \sqrt{7} - \sqrt{11}} \\ = \frac{2 \sqrt{7^{2}} - \sqrt{77} - 4 \sqrt{77} + 2 \sqrt{1 1^{2}}}{{(2 \sqrt{7})}^{2} - {(\sqrt{11})}^{2}} \\ = \frac{2 (7) - 5 \sqrt{77} + 2 (11)}{4 (7) - 11} \\ = \frac{14 - 5 \sqrt{77} + 22}{28 - 11} \\ = \frac{36 - 5 \sqrt{77}}{17} \end{matrix}$

Categories: Capítulo XXXI