▶️ Ejercicio 13 álgebra - Solucionario Baldor ✅✅

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PRELIMINARES

Valor numérico de expresiones compuestas

Ejercicio 13

Para este conjunto de problemas se debe reemplazar cada letra por su respectivo valor numérico y realizar las operaciones correspondientes

Hallar, el valor numérico de las expresiones siguientes para: a=1, b=2, c=3, d=4, m= $\frac{1}{2}$ , n= $\frac{2}{3}$ ,p= $\frac{1}{4}$ , x=0

$(a + b) c - d = (1 + 2) 3 - 4 = 3 (3) - 4 = 9 - 4 = 5$
$(a + b) (b - a) = (1 + 2) (2 - 1) = 3 (1) = 3$
$\begin{matrix} (b - m) (c - n) + 4 a^{2} & = & (2 - \frac{1}{2}) (3 - \frac{2}{3}) + 4 (1)^{2} \\ = & (\frac{4 - 1}{2}) (\frac{9 - 2}{3}) + 4 \\ = & \frac{3}{2} (\frac{7}{3}) + 4 \\ = & \frac{7}{2} + 4 = \frac{7 + 8}{2} = \frac{15}{2} \leftrightarrow 7 \frac{1}{2} \end{matrix}$
$\begin{matrix} (2 m + 3 n) (4 p + b^{2}) & = & (2 \frac{1}{2} + 3 \frac{2}{3}) (4 \frac{1}{4} + 2^{2}) \\ = & (1 + 2) (1 + 4) \\ = & 3 (5) \\ = & 15 \end{matrix}$
$\begin{matrix} (4 m + 8 p) (a^{2} + b^{2}) (6 n - d) & = & (\frac{1}{2} + \frac{1}{4}) (1^{2} + 2^{2}) (\frac{2}{3} - 4) \\ = \\ = & 0 \end{matrix}$
$\begin{matrix} (c - b) (d - c) (b - a) (m - p) & = & (3 - 2) (4 - 3) (2 - 1) (\frac{1}{2} - \frac{1}{4}) \\ = & (1) (1) (1) (\frac{2 - 1}{4}) \\ = & \frac{1}{4} \end{matrix}$
$\begin{matrix} b^{2} (c + d) - a^{2} (m + n) + 2 x & = & 2^{2} (3 + 4) - 1^{2} (\frac{1}{2} + \frac{2}{3}) + 2 (0) \\ = & 4 (7) - (\frac{3 + 4}{6}) \\ = & 28 - \frac{7}{6} = \frac{168 - 7}{6} \\ = & \frac{161}{6} \leftrightarrow 26 \frac{5}{6} \end{matrix}$
$\begin{matrix} 2 m x + 6 (b^{2} + c^{2}) - 4 d^{2} & = & + 6 (2^{2} + 3^{2}) - 4 (4)^{2} \\ = & 6 (13) - 64 \\ = & 78 - 64 \\ = & 14 \end{matrix}$
$\begin{matrix} (\frac{8 m}{9 n} + \frac{16 p}{b}) a & = & (\frac{\frac{1}{2}}{\frac{2}{3}} + \frac{\frac{1}{4}}{2}) (1) \\ = & (\frac{}{} + \frac{}{2}) \\ = & \frac{2}{3} + 2 = \frac{2 + 6}{3} \\ = & \frac{8}{3} \leftrightarrow 2 \frac{2}{3} \end{matrix}$
$\begin{matrix} x + m (a^{b} + d^{c} - c^{a}) & = & 0 + \frac{1}{2} (1^{2} + 4^{3} - 3^{1}) \\ = & \frac{1}{2} (1 + 64 - 3) \\ = & \frac{}{2} \\ = & 31 \end{matrix}$
$\begin{matrix} \frac{4 (m + p)}{a} + \frac{a^{2} + b^{2}}{c^{2}} & = & \frac{4 (\frac{1}{2} + \frac{1}{4})}{1} \div \frac{1^{2} + 2^{2}}{3^{2}} \\ = & 4 (\frac{2 + 1}{4}) \div \frac{5}{9} \\ = & 3 \div \frac{5}{9} \leftrightarrow \frac{3}{\frac{5}{9}} \\ = & \frac{27}{5} \leftrightarrow 5 \frac{2}{5} \end{matrix}$
$\begin{matrix} (2 m + 3 n + 4 p) (8 p + 6 n - 4 m) (9 n + 20 p) & = & (2 \frac{1}{2} + 3 \frac{2}{3} + 4 \frac{1}{4}) (\frac{1}{4} + \frac{2}{3} - \frac{1}{2}) (\frac{2}{3} + \frac{1}{4}) \\ = & (1 + 2 + 1) (2 + 4 - 2) (6 + 5) \\ = & (4) (4) (11) \\ = & 176 \end{matrix}$
$\begin{matrix} c^{2} (m + n) - d^{2} (m + p) + b^{2} (n + p) & = & 3^{2} (\frac{1}{2} + \frac{2}{3}) - 4^{2} (\frac{1}{2} + \frac{1}{4}) + 2^{2} (\frac{2}{3} + \frac{1}{4}) \\ = & (\frac{3 + 4}{}) - (\frac{2 + 1}{4}) + 4 (\frac{8 + 3}{}) \\ = & 3 (\frac{7}{2}) - 4 (3) + \frac{11}{3} \\ = & \frac{21}{2} - 12 + \frac{11}{3} \\ = & \frac{63 - 72 + 22}{6} \\ = & \frac{13}{6} \leftrightarrow 2 \frac{1}{6} \end{matrix}$
$\begin{matrix} (\frac{\sqrt{c^{2} + d^{2}}}{a} \div \frac{2}{\sqrt{d}}) m & = & (\frac{\frac{\sqrt{c^{2} + d^{2}}}{a}}{\frac{2}{\sqrt{d}}}) m \\ = & (\frac{\frac{\sqrt{3^{2} + 4^{2}}}{1}}{\frac{2}{\sqrt{4}}}) \frac{1}{2} \\ = & (\frac{\sqrt{25}}{\frac{2}{2}}) \frac{1}{2} \\ = & \frac{5}{2} \leftrightarrow 2 \frac{1}{2} \end{matrix}$
$\begin{matrix} (4 p + 2 b) (18 n - 24 p) + 2 (8 m + 2) (40 p + a) & = & (4 \frac{1}{4} + 2 (2)) (\frac{2}{3} - \frac{1}{4}) + 2 (\frac{1}{2} + 2) (\frac{1}{4} + 1) \\ = & (1 + 4) (12 - 6) + 2 (4 + 2) (10 + 1) \\ = & 5 (6) + 2 (6) (11) \\ = & 30 + 132 \\ = & 162 \end{matrix}$
$\begin{matrix} \frac{a + \frac{d}{b}}{d - b} \times \frac{5 + \frac{2}{m^{2}}}{p^{2}} & = & \frac{1 + \frac{}{2}}{4 - 2} \times \frac{5 + \frac{2}{{(\frac{1}{2})}^{2}}}{{(\frac{1}{4})}^{2}} \\ = & \frac{3}{2} \times \frac{5 + \frac{2}{\frac{1}{4}}}{\frac{1}{16}} \\ = & \frac{3}{2} \times \frac{5 + 8}{\frac{1}{16}} \\ = & \frac{3}{2} \times \frac{13}{\frac{1}{16}} \\ = & \frac{3}{2} \times 13 ( \\ = & 312 \end{matrix}$
$\begin{matrix} (a + b) \sqrt{c^{2} + 8 b} - m \sqrt{n^{2}} & = & (1 + 2) \sqrt{3^{2} + 8 (2)} - \frac{1}{2} \sqrt[]{{(\frac{2}{3})}^{2}} \\ = & 3 \sqrt{9 + 16} - \frac{1}{2} (\frac{2}{3}) \\ = & 3 \sqrt{25} - \frac{1}{3} \\ = & 3 (5) - \frac{1}{3} \\ = & 15 - \frac{1}{3} = \frac{45 - 1}{3} \\ = & \frac{44}{3} \leftrightarrow 14 \frac{2}{3} \end{matrix}$
$\begin{matrix} (\frac{\sqrt{a + c}}{2} + \frac{\sqrt{6 n}}{b}) \div (c + d) \sqrt{p} & = & (\frac{\sqrt{1 + 3}}{2} + \frac{\sqrt{\frac{2}{3}}}{2}) \div (3 + 4) \sqrt{\frac{1}{4}} \\ = & (\frac{\sqrt{4}}{2} + \frac{\sqrt{4}}{2}) \div 7 (\frac{1}{2}) \\ = & (\frac{2}{2} + \frac{2}{2}) \div \frac{7}{2} \\ = & \frac{2}{\frac{7}{2}} \\ = & \frac{4}{7} \end{matrix}$
$\begin{matrix} 3 (c - b) \sqrt{32 m} - 2 (d - a) \sqrt{16 p} - \frac{2}{n} & = & 3 (3 - 2) \sqrt{\frac{1}{2}} - 2 (4 - 1) \sqrt{\frac{1}{4}} - \frac{2}{\frac{2}{3}} \\ = & 3 (1) (4) - 2 (3) (2) - 3 \\ = & 12 - 12 - 3 \\ = & - 3 \end{matrix}$
$\begin{matrix} \frac{\sqrt{6 a b c}}{2 \sqrt{8 b}} + \frac{\sqrt{3 m n}}{2 (b - a)} - \frac{c d n p}{a b c} & = & \frac{\sqrt{6 (1) (2) (3)}}{2 \sqrt{8 (2)}} + \frac{\sqrt{3 (\frac{1}{2}) (\frac{2}{3})}}{2 (2 - 1)} - \frac{3 (4) (\frac{2}{3}) (\frac{1}{4})}{(1) (2) (3)} \\ = & \frac{\sqrt{36}}{2 \sqrt{16}} + \frac{1}{2} - \frac{1}{3} \\ = & \frac{}{2 (4)} + (\frac{3 - 2}{6}) \\ = & \frac{3}{4} + \frac{1}{6} = \frac{9 + 2}{12} \\ = & \frac{11}{12} \end{matrix}$
$\begin{matrix} \frac{a^{2} + b^{2}}{b^{2} - a^{2}} + 3 (a + b) (2 a + 3 b) & = & \frac{1^{2} + 2^{2}}{2^{2} - 1^{2}} + 3 (1 + 2) (2 \times 1 + 3 \times 2) \\ = & \frac{1 + 4}{4 - 1} + 3 (3) (2 + 6) \\ = & \frac{5}{3} + 72 = \frac{5 + 216}{3} \\ = & \frac{221}{3} \leftrightarrow 73 \frac{2}{3} \end{matrix}$
$\begin{matrix} b^{2} + (\frac{1}{a} + \frac{1}{b}) (\frac{1}{b} + \frac{1}{c}) + {(\frac{1}{n} + \frac{1}{m})}^{2} & = & 2^{2} + (\frac{1}{1} + \frac{1}{2}) (\frac{1}{2} + \frac{1}{3}) + {(\frac{1}{\frac{2}{3}} + \frac{1}{\frac{1}{2}})}^{2} \\ = & 4 + (\frac{2 + 1}{2}) (\frac{3 + 2}{6}) + {(\frac{3}{2} + 2)}^{2} \\ = & 4 + (\frac{3}{2}) (\frac{5}{}) + {(\frac{3 + 4}{2})}^{2} \\ = & 4 + \frac{5}{4} + {(\frac{7}{2})}^{2} \\ = & (\frac{16 + 5}{4}) + \frac{49}{4} \\ = & \frac{21}{4} + \frac{49}{4} = \frac{21 + 49}{4} \\ \frac{}{} \\ = & \frac{35}{2} \leftrightarrow 17 \frac{1}{2} \end{matrix}$
$\begin{matrix} (2 m + 3 n) (4 p + 2 c) - 4 m^{2} n^{2} & = & (2 \frac{1}{2} + 3 \frac{2}{3}) (4 \frac{1}{4} + 2 \times 3) - 4 {(\frac{1}{2})}^{2} {(\frac{2}{3})}^{2} \\ = & (1 + 2) (1 + 6) - 4 (\frac{1}{4}) (\frac{4}{9}) \\ = & 3 (7) - \frac{4}{9} \\ = & 21 - \frac{4}{9} = \frac{189 - 4}{9} \\ = & \frac{185}{9} \leftrightarrow 20 \frac{5}{9} \end{matrix}$
$\begin{matrix} \frac{b^{2} - \frac{c}{3}}{2 a b - m} - \frac{n}{b - m} & = & \frac{2^{2} - \frac{3}{3}}{2 (1) (2) - \frac{1}{2}} - \frac{\frac{2}{3}}{2 - \frac{1}{2}} \\ = & \frac{4 - 1}{4 - \frac{1}{2}} - \frac{\frac{2}{3}}{\frac{4 - 1}{2}} \\ = & \frac{3}{\frac{8 - 1}{2}} - \frac{\frac{2}{3}}{\frac{3}{2}} \\ = & \frac{3}{\frac{7}{2}} - \frac{4}{9} \\ = & \frac{6}{7} - \frac{4}{9} = \frac{54 - 28}{63} \\ = & \frac{26}{63} \end{matrix}$