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CAPITULO XXXI

R a d i c a l e s
Ejercicio 248
1. $\begin{array}{cc}\frac{3–\sqrt{2}}{1+\sqrt{2}}& =\frac{3–\sqrt{2}}{1+\sqrt{2}}×\frac{1–\sqrt{2}}{1–\sqrt{2}}\\ & =\frac{3–\sqrt{2}–3\sqrt{2}+\sqrt{{2}^{2}}}{1–\sqrt{{2}^{2}}}\\ & =\frac{3–4\sqrt{2}+2}{1–2}\\ & =\frac{5–4\sqrt{2}}{–1}\\ & =4\sqrt{2}–5\end{array}$
2. $\begin{array}{cc}\frac{5+2\sqrt{3}}{4–\sqrt{3}}& =\frac{5+2\sqrt{3}}{4–\sqrt{3}}×\frac{4+\sqrt{3}}{4+\sqrt{3}}\\ & =\frac{20+8\sqrt{3}+5\sqrt{3}+2\sqrt{{3}^{2}}}{{4}^{2}–\sqrt{{3}^{2}}}\\ & =\frac{20+13\sqrt{3}+2\left(3\right)}{16–3}\\ & =\frac{13\sqrt{3}+26}{13}\\ & =\frac{\overline{)13}\left(\sqrt{3}+2\right)}{\overline{)13}}\\ & =\sqrt{3}+2\end{array}$
3. $\begin{array}{cc}\frac{\sqrt{2}–\sqrt{5}}{\sqrt{2}+\sqrt{5}}& =\frac{\sqrt{2}–\sqrt{5}}{\sqrt{2}+\sqrt{5}}×\frac{\sqrt{2}–\sqrt{5}}{\sqrt{2}–\sqrt{5}}\\ & =\frac{{\left(\sqrt{2}–\sqrt{5}\right)}^{2}}{\sqrt{{2}^{2}}–\sqrt{{5}^{2}}}\\ & =\frac{\sqrt{{2}^{2}}–2\sqrt{2×5}+\sqrt{{5}^{2}}}{2–5}\\ & =\frac{2–2\sqrt{10}+5}{–3}\\ & =\frac{7–2\sqrt{10}}{–3}\\ & =\frac{2\sqrt{10}–7}{3}\end{array}$
4. $\begin{array}{cc}\frac{\sqrt{7}+2\sqrt{5}}{\sqrt{7}–\sqrt{5}}& =\frac{\sqrt{7}+2\sqrt{5}}{\sqrt{7}–\sqrt{5}}×\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\ & =\frac{\sqrt{{7}^{2}}+2\sqrt{35}+\sqrt{35}+2\sqrt{{5}^{2}}}{\sqrt{{7}^{2}}–\sqrt{{5}^{2}}}\\ & =\frac{7+3\sqrt{35}+2\left(5\right)}{7–5}\\ & =\frac{17+3\sqrt{35}}{2}\end{array}$
5. $\begin{array}{cc}\frac{\sqrt{2}–3\sqrt{5}}{2\sqrt{2}+\sqrt{5}}& =\frac{\sqrt{2}–3\sqrt{5}}{2\sqrt{2}+\sqrt{5}}×\frac{2\sqrt{2}–\sqrt{5}}{2\sqrt{2}–\sqrt{5}}\\ & =\frac{2\sqrt{{2}^{2}}–6\sqrt{10}–\sqrt{10}+3\sqrt{{5}^{2}}}{{\left(2\sqrt{2}\right)}^{2}–\sqrt{{5}^{2}}}\\ & =\frac{2\left(2\right)–7\sqrt{10}+3\left(5\right)}{4\left(2\right)–5}\\ & =\frac{4–7\sqrt{10}+15}{8–5}\\ & =\frac{19–7\sqrt{10}}{3}\end{array}$
6. $\begin{array}{cc}\frac{19}{5\sqrt{2}–4\sqrt{3}}& =\frac{19}{5\sqrt{2}–4\sqrt{3}}×\frac{5\sqrt{2}+4\sqrt{3}}{5\sqrt{2}+4\sqrt{3}}\\ & =\frac{19\left(5\sqrt{2}+4\sqrt{3}\right)}{{\left(5\sqrt{2}\right)}^{2}–{\left(4\sqrt{3}\right)}^{2}}\\ & =\frac{19\left(5\sqrt{2}+4\sqrt{3}\right)}{25\left(2\right)–16\left(3\right)}\\ & =\frac{19\left(5\sqrt{2}+4\sqrt{3}\right)}{50–48}\\ & =\frac{19\left(5\sqrt{2}+4\sqrt{3}\right)}{2}\end{array}$
7. $\begin{array}{cc}\frac{3\sqrt{2}}{7\sqrt{2}–6\sqrt{3}}& =\frac{3\sqrt{2}}{7\sqrt{2}–6\sqrt{3}}×\frac{7\sqrt{2}+6\sqrt{3}}{7\sqrt{2}+6\sqrt{3}}\\ & =\frac{21\sqrt{{2}^{2}}+18\sqrt{6}}{{\left(7\sqrt{2}\right)}^{2}–{\left(6\sqrt{3}\right)}^{2}}\\ & =\frac{21\left(2\right)+18\sqrt{6}}{49\left(2\right)–36\left(3\right)}\\ & =\frac{42+18\sqrt{6}}{98–108}\\ & =\frac{\overline{)2}\left(21+9\sqrt{6}\right)}{–}\\ & =–\frac{21+9\sqrt{6}}{5}\end{array}$
8. $\begin{array}{cc}\frac{4\sqrt{3}–3\sqrt{7}}{2\sqrt{3}+3\sqrt{7}}& =\frac{4\sqrt{3}–3\sqrt{7}}{2\sqrt{3}+3\sqrt{7}}×\frac{2\sqrt{3}–3\sqrt{7}}{2\sqrt{3}–3\sqrt{7}}\\ & =\frac{8\sqrt{{3}^{2}}–6\sqrt{21}–12\sqrt{21}+9\sqrt{{7}^{2}}}{{\left(2\sqrt{3}\right)}^{2}–{\left(3\sqrt{7}\right)}^{2}}\\ & =\frac{8\left(3\right)–18\sqrt{21}+9\left(7\right)}{4\left(3\right)–9\left(7\right)}\\ & =\frac{24–18\sqrt{21}+63}{12–63}\\ & =\frac{87–18\sqrt{21}}{–51}\\ & =\frac{18\sqrt{21}–87}{51}\\ & =\frac{\overline{)3}\left(6\sqrt{21}–29\right)}{}\\ & =\frac{6\sqrt{21}–29}{17}\end{array}$
9. $\begin{array}{cc}\frac{5\sqrt{2}–6\sqrt{3}}{4\sqrt{2}–3\sqrt{3}}& =\frac{5\sqrt{2}–6\sqrt{3}}{4\sqrt{2}–3\sqrt{3}}×\frac{4\sqrt{2}+3\sqrt{3}}{4\sqrt{2}+3\sqrt{3}}\\ & =\frac{20\sqrt{{2}^{2}}–24\sqrt{6}+15\sqrt{6}–18\sqrt{{3}^{2}}}{{\left(4\sqrt{2}\right)}^{2}–{\left(3\sqrt{3}\right)}^{2}}\\ & =\frac{20\left(2\right)–9\sqrt{6}–18\left(3\right)}{16\left(2\right)–9\left(3\right)}\\ & =\frac{40–9\sqrt{6}–54}{32–27}\\ & =\frac{40–9\sqrt{6}–54}{32–27}\\ & =–\frac{14+9\sqrt{6}}{5}\end{array}$
10. $\begin{array}{cc}\frac{\sqrt{7}+3\sqrt{11}}{5\sqrt{7}+4\sqrt{11}}& =\frac{\sqrt{7}+3\sqrt{11}}{5\sqrt{7}+4\sqrt{11}}×\frac{5\sqrt{7}–4\sqrt{11}}{5\sqrt{7}–4\sqrt{11}}\\ & =\frac{5\sqrt{{7}^{2}}+15\sqrt{77}–4\sqrt{77}–12\sqrt{1{1}^{2}}}{{\left(5\sqrt{7}\right)}^{2}–{\left(4\sqrt{11}\right)}^{2}}\\ & =\frac{5\left(7\right)+11\sqrt{77}–12\left(11\right)}{25\left(7\right)–16\left(11\right)}\\ & =\frac{35+11\sqrt{77}–132}{175–176}\\ & =97–11\sqrt{77}\end{array}$
11. $\begin{array}{cc}\frac{\sqrt{5}+\sqrt{2}}{7+2\sqrt{10}}& =\frac{\sqrt{5}+\sqrt{2}}{7+2\sqrt{10}}×\frac{7–2\sqrt{10}}{7–2\sqrt{10}}\\ & =\frac{7\sqrt{5}+7\sqrt{2}–2\sqrt{50}–2\sqrt{20}}{{7}^{2}–{\left(2\sqrt{10}\right)}^{2}}\\ & =\frac{7\sqrt{5}+7\sqrt{2}–2\sqrt{{5}^{2}.2}–2\sqrt{5.{2}^{2}}}{49–4\left(10\right)}\\ & =\frac{7\sqrt{5}+7\sqrt{2}–10\sqrt{2}–4\sqrt{5}}{9}\\ & =\frac{3\sqrt{5}–3\sqrt{2}}{9}\\ & =\frac{\overline{)3}\left(\sqrt{5}–\sqrt{2}\right)}{}\\ & =\frac{\sqrt{5}–\sqrt{2}}{3}\end{array}$
12. $\begin{array}{cc}\frac{9\sqrt{3}–3\sqrt{2}}{6–\sqrt{6}}& =\frac{9\sqrt{3}–3\sqrt{2}}{6–\sqrt{6}}×\frac{6+\sqrt{6}}{6+\sqrt{6}}\\ & =\frac{54\sqrt{3}–18\sqrt{2}+9\sqrt{18}–3\sqrt{12}}{{6}^{2}–\sqrt{{6}^{2}}}\\ & =\frac{54\sqrt{3}–18\sqrt{2}+9\sqrt{{3}^{2}.2}–3\sqrt{3.{2}^{2}}}{36–6}\\ & =\frac{54\sqrt{3}–18\sqrt{2}+27\sqrt{2}–6\sqrt{3}}{30}\\ & =\frac{48\sqrt{3}+9\sqrt{2}}{30}\\ & =\frac{\overline{)3}\left(16\sqrt{3}+3\sqrt{2}\right)}{}\\ & =\frac{16\sqrt{3}+3\sqrt{2}}{10}\end{array}$
13. $\begin{array}{cc}\frac{\sqrt{a}+\sqrt{x}}{2\sqrt{a}+\sqrt{x}}& =\frac{\sqrt{a}+\sqrt{x}}{2\sqrt{a}+\sqrt{x}}×\frac{2\sqrt{a}–\sqrt{x}}{2\sqrt{a}–\sqrt{x}}\\ & =\frac{2\sqrt{{a}^{2}}+2\sqrt{ax}–\sqrt{ax}–\sqrt{{x}^{2}}}{{\left(2\sqrt{a}\right)}^{2}–{\left(\sqrt{x}\right)}^{2}}\\ & =\frac{2a+\sqrt{ax}–x}{4a–x}\end{array}$
14. $\begin{array}{cc}\frac{\sqrt{x}–\sqrt{x–1}}{\sqrt{x}+\sqrt{x–1}}& =\frac{\sqrt{x}–\sqrt{x–1}}{\sqrt{x}+\sqrt{x–1}}×\frac{\sqrt{x}–\sqrt{x–1}}{\sqrt{x}–\sqrt{x–1}}\\ & =\frac{\sqrt{{x}^{2}}–\sqrt{x\left(x–1\right)}–\sqrt{x\left(x–1\right)}+\sqrt{{\left(x–1\right)}^{2}}}{\sqrt{{x}^{2}}–\sqrt{{\left(x–1\right)}^{2}}}\\ & =\frac{x–2\sqrt{x\left(x–1\right)}+x–1}{x–\left(x–1\right)}\\ & =\frac{2x–2\sqrt{x\left(x–1\right)}–1}{\overline{)x}–\overline{)x}+1}\\ & =2x–2\sqrt{x\left(x–1\right)}–1\end{array}$
15. $\begin{array}{cc}\frac{\sqrt{a}–\sqrt{a+1}}{\sqrt{a}+\sqrt{a+1}}& =\frac{\sqrt{a}–\sqrt{a+1}}{\sqrt{a}+\sqrt{a+1}}×\frac{\sqrt{a}–\sqrt{a+1}}{\sqrt{a}–\sqrt{a+1}}\\ & =\frac{{\left(\sqrt{a}–\sqrt{a+1}\right)}^{2}}{{\left(\sqrt{a}\right)}^{2}–{\left(\sqrt{a+1}\right)}^{2}}\\ & =\frac{a–2\sqrt{a\left(a+1\right)}+a+1}{a–\left(a+1\right)}\\ & =\frac{2a–2\sqrt{a\left(a+1\right)}+1}{\overline{)a}–\overline{)a}–1}\\ & =2\sqrt{a\left(a+1\right)}–2a–1\end{array}$
16. $\begin{array}{cc}\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}–\sqrt{2}}& =\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}–\sqrt{2}}×\frac{\sqrt{x+2}+\sqrt{2}}{\sqrt{x+2}+\sqrt{2}}\\ & =\frac{{\left(\sqrt{x+2}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{x+2}\right)}^{2}–{\left(\sqrt{2}\right)}^{2}}\\ & =\frac{x+2+2\sqrt{2\left(x+2\right)}+2}{x+\overline{)2}–\overline{)2}}\\ & =\frac{x+2\sqrt{2\left(x+2\right)}+4}{x}\end{array}$
17. $\begin{array}{cc}\frac{\sqrt{a+4}–\sqrt{a}}{\sqrt{a+4}+\sqrt{a}}& =\frac{\sqrt{a+4}–\sqrt{a}}{\sqrt{a+4}+\sqrt{a}}×\frac{\sqrt{a+4}–\sqrt{a}}{\sqrt{a+4}–\sqrt{a}}\\ & =\frac{{\left(\sqrt{a+4}–\sqrt{a}\right)}^{2}}{{\left(\sqrt{a+4}\right)}^{2}–{\left(\sqrt{a}\right)}^{2}}\\ & =\frac{a+4–2\sqrt{a\left(a+4\right)}+a}{\overline{)a}+4–\overline{)a}}\\ & =\frac{2a–2\sqrt{a\left(a+4\right)}+4}{4}\\ & =\frac{\overline{)2}\left(a–\sqrt{a\left(a+4\right)}+2\right)}{}\\ & =\frac{a–\sqrt{a\left(a+4\right)}+2}{2}\end{array}$
18. $\begin{array}{cc}\frac{\sqrt{a+b}–\sqrt{a–b}}{\sqrt{a+b}+\sqrt{a–b}}& =\frac{\sqrt{a+b}–\sqrt{a–b}}{\sqrt{a+b}+\sqrt{a–b}}×\frac{\sqrt{a+b}–\sqrt{a–b}}{\sqrt{a+b}–\sqrt{a–b}}\\ & =\frac{{\left(\sqrt{a+b}–\sqrt{a–b}\right)}^{2}}{{\left(\sqrt{a+b}\right)}^{2}–{\left(\sqrt{a–b}\right)}^{2}}\\ & =\frac{a+\overline{)b}–2\sqrt{{a}^{2}–{b}^{2}}+a–\overline{)b}}{a+b–\left(a–b\right)}\\ & =\frac{2a–2\sqrt{{a}^{2}–{b}^{2}}}{\overline{)a}+b–\overline{)a}+b}\\ & =\frac{\overline{)2}\left(a–\sqrt{{a}^{2}–{b}^{2}}\right)}{\overline{)2}b}\\ & =\frac{a–\sqrt{{a}^{2}–{b}^{2}}}{b}\end{array}$