▶️ Ejercicio 222 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXX

TEORIA DE LOS EXPONENTES

Ejercicio 222

Hallar el valor numérico de:

$\begin{matrix} a^{- 2} + a^{- 1} b^{\frac{1}{2}} + x^{0} p a r a a & = 3, b = 4 \\ a^{- 2} + a^{- 1} b^{\frac{1}{2}} + & = \frac{1}{a^{2}} + \frac{b^{\frac{1}{2}}}{a} + 1 \\ = \frac{1}{3^{2}} + \frac{4^{\frac{1}{2}}}{3} + 1 \\ = \frac{1}{9} + \frac{2}{3} + 1 \\ = \frac{1 + 6 + 9}{9} \\ = \frac{16}{9} \leftrightarrow 1 \frac{7}{9} \end{matrix}$
$\begin{matrix} 3 x^{- \frac{1}{2}} + x^{2} y^{- 3} + x^{0} y^{\frac{1}{3}} p a r a x & = 4, y = 1 \\ 3 x^{- \frac{1}{2}} + x^{2} y^{- 3} + y^{\frac{1}{3}} & = \frac{3}{x^{\frac{1}{2}}} + \frac{x^{2}}{y^{3}} + y^{\frac{1}{3}} \\ = \frac{3}{4^{\frac{1}{2}}} + \frac{4^{2}}{1^{3}} + 1^{\frac{1}{3}} \\ = \frac{3}{2} + 16 + 1 \\ = \frac{3 + 32 + 2}{2} \\ = \frac{37}{2} \\ = 18 \frac{1}{2} \end{matrix}$
$\begin{matrix} 2 a^{- 3} b + \frac{a^{- 4}}{b^{- 1}} + a^{\frac{1}{2}} b^{- \frac{3}{4}} p a r a a & = 4, b = 16 \\ 2 a^{- 3} b + \frac{a^{- 4}}{b^{- 1}} + a^{\frac{1}{2}} b^{- \frac{3}{4}} & = \frac{2 b}{a^{3}} + \frac{b}{a^{4}} + \frac{a^{\frac{1}{2}}}{b^{\frac{3}{4}}} \\ = \frac{2 (16)}{4^{3}} + \frac{16}{4^{4}} + \frac{4^{\frac{1}{2}}}{1 6^{\frac{3}{4}}} \\ = \frac{2 (4^{2})}{4^{3}} + \frac{4^{2}}{4^{}} + \frac{4^{\frac{1}{2}}}{4^{\frac{3}{2}}} \\ = \frac{1}{2} + \frac{1}{16} + \frac{1}{4} \\ = \frac{8 + 1 + 4}{16} \\ = \frac{13}{16} \end{matrix}$
$\begin{matrix} \frac{x^{\frac{3}{4}}}{y^{- 2}} + x^{- \frac{1}{2}} y^{- \frac{1}{3}} + x^{0} y^{0} + \frac{x}{y^{\frac{4}{3}}} p a r a x & = 16, y = 8 \\ \frac{x^{\frac{3}{4}}}{y^{- 2}} + x^{- \frac{1}{2}} y^{- \frac{1}{3}} - + \frac{x}{y^{\frac{4}{3}}} & = x^{\frac{3}{4}} y^{2} + \frac{1}{x^{\frac{1}{2}} y^{\frac{1}{3}}} - 1 + \frac{x}{y^{\frac{4}{3}}} \\ = 1 6^{\frac{3}{4}} 8^{2} + \frac{1}{1 6^{\frac{1}{2}} 8^{\frac{1}{3}}} - 1 + \frac{16}{8^{\frac{4}{3}}} \\ = {(2^{4})}^{\frac{3}{4}} . {(2^{3})}^{2} + \frac{1}{{(2^{4})}^{\frac{1}{2}} . {(2^{3})}^{\frac{1}{3}}} - 1 + \frac{2^{4}}{{(2^{3})}^{\frac{4}{3}}} \\ = 2^{3} . 2^{6} + \frac{1}{2^{2} .2} - 1 + \frac{2^{4}}{2^{4}} \\ = 2^{9} + \frac{1}{2^{3}} - 1 + 1 \\ = \frac{2^{9 + 3} + 1}{2^{3}} \\ = \frac{2^{12} + 1}{2^{3}} \\ = \frac{4096 + 1}{8} \\ = \frac{4097}{8} \\ = 512 \frac{1}{8} \end{matrix}$
$\begin{matrix} \frac{x^{0}}{x^{- 1}} + \frac{y^{- 3}}{y^{0}} + 2 x^{0} + x^{\frac{3}{4}} y^{2} p a r a x & = 81, y = 3 \\ \frac{}{x^{- 1}} + \frac{y^{- 3}}{} + 2 + x^{\frac{3}{4}} y^{- 2} & = x + \frac{1}{y^{3}} + 2 + x^{\frac{3}{4}} y^{- 2} \\ = 81 + \frac{1}{3^{3}} + 2 + 8 1^{\frac{3}{4}} 3^{- 2} \\ = 3^{4} + \frac{1}{3^{3}} + 2 + {(3^{4})}^{\frac{3}{4}} 3^{- 2} \\ = 3^{4} + \frac{1}{3^{3}} + 2 + 3^{3} . 3^{- 2} \\ = 3^{4} + \frac{1}{3^{3}} + 2 + 3 \\ = \frac{3^{7} + 1 + 5. 3^{3}}{3^{3}} \\ = \frac{2187 + 1 + 135}{27} \\ = \frac{2323}{27} \leftrightarrow 86 \frac{1}{27} \end{matrix}$
$\begin{matrix} a^{\frac{1}{2}} x^{\frac{1}{3}} + a^{- \frac{1}{2}} x^{- \frac{1}{3}} + \frac{1}{a^{- \frac{1}{4}} x^{- 1}} + 3 x^{0} p a r a a & = 16, x = 8 \\ a^{\frac{1}{2}} x^{\frac{1}{3}} + a^{- \frac{1}{2}} x^{- \frac{1}{3}} + \frac{1}{a^{- \frac{1}{4}} x^{- 1}} + 3 & = a^{\frac{1}{2}} x^{\frac{1}{3}} + \frac{1}{a^{\frac{1}{2}} x^{\frac{1}{3}}} + a^{\frac{1}{4}} x + 3 \\ = 1 6^{\frac{1}{2}} 8^{\frac{1}{3}} + \frac{1}{1 6^{\frac{1}{2}} 8^{\frac{1}{3}}} + 1 6^{\frac{1}{4}} 8 + 3 \\ = {(2^{4})}^{\frac{1}{2}} {(2^{3})}^{\frac{1}{3}} + \frac{1}{{(2^{4})}^{\frac{1}{2}} {(2^{3})}^{\frac{1}{3}}} + {(2^{4})}^{\frac{1}{4}} (2^{3}) + 3 \\ = 2^{2} .2 + \frac{1}{2^{2} .2} + 2. 2^{3} + 3 \\ = 2^{3} + \frac{1}{2^{3}} + 2^{4} + 3 \\ = \frac{2^{6} + 1 + 2^{7} + 3. 2^{3}}{2^{3}} \\ = \frac{64 + 1 + 128 + 24}{8} \\ = \frac{217}{8} \leftrightarrow 27 \frac{1}{8} \end{matrix}$
$\begin{matrix} \frac{a^{- 3}}{b^{- 1}} + 3 a^{- 1} b^{2} c^{- 3} - \frac{a^{- 2}}{b^{\frac{1}{2}} c^{- 1}} + b^{\frac{1}{4}} + c^{0} p a r a a & = 3, b = 16, c = 2 \\ \frac{a^{- 2}}{b^{- 1}} + 3 a^{- 1} b^{2} c^{- 3} - \frac{a^{- 2}}{b^{\frac{1}{2}} c^{- 1}} + b^{\frac{1}{4}} + & = \frac{b}{a^{2}} + \frac{3 b^{2}}{a c^{3}} - \frac{c}{a^{2} b^{\frac{1}{2}}} + b^{\frac{1}{4}} + 1 \\ = \frac{16}{3^{2}} + \frac{3 {(16)}^{2}}{3 {(2)}^{3}} - \frac{2}{3^{2} {(16)}^{\frac{1}{2}}} + 1 6^{\frac{1}{4}} + 1 \\ = \frac{2^{4}}{3^{2}} + \frac{{(2^{4})}^{2}}{2^{3}} - \frac{2}{3^{2} {(2^{4})}^{\frac{1}{2}}} + {(2^{4})}^{\frac{1}{4}} + 1 \\ = \frac{2^{4}}{3^{2}} + \frac{2^{}}{2^{3}} - \frac{2}{3^{2} (2^{2})} + 2 + 1 \\ = \frac{2^{4}}{3^{2}} + 2^{5} - \frac{1}{3^{2} .2} + 3 \\ = \frac{2^{5} + 3^{2} . 2^{6} - 1 + 3^{3} .2}{3^{2} .2} \\ = \frac{32 + 576 - 1 + 54}{18} \\ = \frac{661}{18} \leftrightarrow 36 \frac{13}{18} \end{matrix}$
$\begin{matrix} \frac{x^{0}}{3 y^{0}} + x^{\frac{2}{3}} - y^{\frac{1}{5}} + \frac{x^{- 2}}{y^{- 1}} + y^{0} p a r a x & = 8, y = 32 \\ \frac{}{3} + x^{\frac{2}{3}} - y^{\frac{1}{5}} + \frac{x^{- 2}}{y^{- 1}} + & = \frac{1}{3} + x^{\frac{2}{3}} - y^{\frac{1}{5}} + \frac{y}{x^{2}} + 1 \\ = \frac{1}{3} + 8^{\frac{2}{3}} - 3 2^{\frac{1}{5}} + \frac{32}{8^{2}} + 1 \\ = \frac{4}{3} + {(2^{3})}^{\frac{2}{3}} - {(2^{5})}^{\frac{1}{5}} + \frac{2^{5}}{{(2^{3})}^{2}} \\ = \frac{4}{3} + 2^{2} - 2 + \frac{2^{5}}{2^{6}} \\ = \frac{4}{3} + 2 + \frac{1}{2} \\ = \frac{8 + 12 + 3}{6} \\ = \frac{23}{6} \leftrightarrow 3 \frac{5}{6} \end{matrix}$
$\begin{matrix} a^{- \frac{1}{3}} - \frac{1}{b^{- \frac{4}{5}}} + a^{0} b - \sqrt[3]{a} b^{\frac{2}{5}} - \frac{1}{a^{- \frac{2}{3}}} p a r a a & = 27, b = 243 \\ a^{- \frac{1}{3}} - \frac{1}{b^{- \frac{4}{5}}} + a^{0} b - \sqrt[3]{a} b^{\frac{2}{5}} - \frac{1}{a^{- \frac{2}{3}}} & = \frac{1}{a^{\frac{1}{3}}} - b^{\frac{4}{5}} + b - \sqrt[3]{a} b^{\frac{2}{5}} - a^{\frac{2}{3}} \\ = \frac{1}{2 7^{\frac{1}{3}}} - 24 3^{\frac{4}{5}} + 243 - \sqrt[3]{27} 24 3^{\frac{2}{5}} - 2 7^{\frac{2}{3}} \\ = \frac{1}{{(3^{3})}^{\frac{1}{3}}} - {(3^{5})}^{\frac{4}{5}} + 3^{5} - \sqrt[3]{3^{3}} . {(3^{5})}^{\frac{2}{5}} - {(3^{3})}^{\frac{2}{3}} \\ = \frac{1}{3} - 3^{4} + 3^{5} - 3. 3^{2} - 3^{2} \\ = \frac{1 - 3^{5} + 3^{6} - 3^{4} - 3^{3}}{3} \\ = \frac{379}{3} \leftrightarrow 126 \frac{1}{3} \end{matrix}$

Categories: Capítulo XXX