Ejercicio 139 – Solucionario Baldor

Comparte esto 👍👍

DESCARGA

CAPITULO XIV

Operaciones con Fracciones

Ejercicio 139

Hallar el verdadero valor de:

$\begin{matrix} \frac{x - 2}{x + 3} p a r a x & = 2 \\ \frac{x - 2}{x + 3} & = \frac{2 - 2}{2 + 3} \\ = \frac{0}{5} \\ = 0 \end{matrix}$
$\begin{matrix} \frac{x - 2}{x - 3} p a r a x & = 3 \\ \frac{x - 2}{x - 3} & = \frac{3 - 2}{3 - 3} \\ = \frac{1}{0} \\ = \infty \end{matrix}$
$\begin{matrix} \frac{x^{2} - a^{2}}{x^{2} + a^{2}} p a r a x & = a \\ \frac{x^{2} - a^{2}}{x^{2} + a^{2}} & = \frac{a^{2} - a^{2}}{a^{2} + a^{2}} \\ = \frac{0}{2 a^{2}} \end{matrix}$
$\begin{matrix} \frac{x^{2} + y^{2}}{x^{2} - y^{2}} p a r a x & = y \\ \frac{x^{2} + y^{2}}{x^{2} - y^{2}} & = \frac{y^{2} + y^{2}}{y^{2} - y^{2}} \\ = \frac{2 y^{2}}{0} \\ = \infty \end{matrix}$
$\begin{matrix} \frac{x - 1}{\frac{3}{x - 2}} p a r a x & = 2 \\ \frac{x - 1}{\frac{3}{x - 2}} & = \frac{(x - 1) (x - 2)}{3} \\ = \frac{x^{2} - 3 x + 2}{3} \\ = \frac{2^{2} - 3 (2) + 2}{3} \\ = \frac{4 - 6 + 2}{3} \\ = \frac{0}{3} \\ = 0 \end{matrix}$
$\begin{matrix} \frac{x^{2} - 9}{x^{2} + x - 12} p a r a x & = 3 \\ \frac{x^{2} - 9}{x^{2} + x - 12} & = \frac{(x + 3) (x - 3)}{(x + 4) (x - 3)} \\ = \frac{x + 3}{x + 4} \\ = \frac{3 + 3}{3 + 4} \\ = \frac{6}{7} \end{matrix}$
$\begin{matrix} \frac{a^{2} - a - 6}{a^{2} + 2 a - 15} p a r a a & = 3 \\ \frac{a^{2} - a - 6}{a^{2} + 2 a - 15} & = \frac{(a - 3) (a + 2)}{(a + 5) (a - 3)} \\ = \frac{a + 2}{a + 5} \\ = \frac{3 + 2}{3 + 5} \\ = \frac{5}{8} \end{matrix}$
$\begin{matrix} \frac{x^{2} - 7 x + 10}{x^{3} - 2 x^{2} - x + 2} p a r a x & = 2 \\ \frac{x^{2} - 7 x + 10}{x^{3} - 2 x^{2} - x + 2} & = \frac{(x - 5) (x - 2)}{x^{2} (x - 2) - (x - 2)} \\ = \frac{(x - 5) (x - 2)}{(x^{2} - 1) (x - 2)} \\ = \frac{x - 5}{x^{2} - 1} \\ = \frac{2 - 5}{2^{2} - 1} \\ = - \frac{3}{3} \\ = - 1 \end{matrix}$
$\begin{matrix} \frac{x^{2} - 2 x + 1}{x^{3} - 2 x^{2} - x + 2} p a r a x & = 1 \\ \frac{x^{2} - 2 x + 1}{x^{3} - 2 x^{2} - x + 2} & = \frac{{(x - 1)}^{2}}{x^{2} (x - 2) - (x - 2)} \\ = \frac{{(x - 1)}^{2}}{(x^{2} - 1) (x - 2)} \\ = \frac{{(x - 1)}^{2}}{(x + 1) (x - 1) (x - 2)} \\ = \frac{x - 1}{(x + 1) (x - 2)} \\ = \frac{1 - 1}{(1 + 1) (1 - 2)} \\ = \frac{0}{2 (- 1)} \\ = 0 \end{matrix}$
$\begin{matrix} \frac{a^{3} - 8}{a^{2} + 11 a - 26} p a r a a & = 2 \\ \frac{a^{3} - 8}{a^{2} + 11 a - 26} & = \frac{(a - 2) (a^{2} + 2 a + 4)}{(a + 13) (a - 2)} \\ = \frac{a^{2} + 2 a + 4}{a + 13} \\ = \frac{2^{2} + 2 (2) + 4}{2 + 13} \\ = \frac{4 + 4 + 4}{2 + 13} \\ = \frac{}{} \\ = \frac{4}{5} \end{matrix}$
$\begin{matrix} \frac{x^{2} - 7 x + 6}{x^{2} - 2 x + 1} p a r a x & = 1 \\ \frac{x^{2} - 7 x + 6}{x^{2} - 2 x + 1} & = \frac{(x - 6) (x - 1)}{{(x - 1)}^{2}} \\ = \frac{x - 6}{x - 1} \\ = \frac{1 - 6}{1 - 1} \\ = \frac{- 5}{0} \\ = \infty \end{matrix}$
$\begin{matrix} \frac{x^{2} - 16}{x^{3} - 4 x^{2} - x + 4} p a r a x & = 4 \\ \frac{x^{2} - 16}{x^{3} - 4 x^{2} - x + 4} & = \frac{(x - 4) (x + 4)}{x^{2} (x - 4) - (x - 4)} \\ = \frac{(x - 4) (x + 4)}{(x^{2} - 1) (x - 4)} \\ = \frac{x + 4}{(x - 1) (x + 1)} \\ = \frac{4 + 4}{(4 - 1) (4 + 1)} \\ = \frac{8}{3 (5)} \\ = \frac{8}{15} \end{matrix}$
$\begin{matrix} \frac{4 x^{2} - 4 x + 1}{4 x^{2} + 8 x - 5} p a r a x & = \frac{1}{2} \\ \frac{4 x^{2} - 4 x + 1}{4 x^{2} + 8 x - 5} & = \frac{{(2 x - 1)}^{2}}{4 x^{2} - 2 x + 10 x - 5} \\ = \frac{{(2 x - 1)}^{2}}{2 x (2 x - 1) + 5 (2 x - 1)} \\ = \frac{{(2 x - 1)}^{2}}{(2 x + 5) (2 x - 1)} \\ = \frac{2 x - 1}{2 x + 5} \\ = \frac{2 (\frac{1}{2}) - 1}{2 (\frac{1}{2}) + 5} \\ = \frac{1 - 1}{1 + 5} \\ = 0 \end{matrix}$
$\begin{matrix} \frac{x^{3} - a^{3}}{x - a} p a r a x & = a \\ \frac{x^{3} - a^{3}}{x - a} & = \frac{(x - a) (x^{2} + a x + a^{2})}{(x - a)} \\ = x^{2} + a x + a^{2} \\ = a^{2} + a (a) + a^{2} \\ = 3 a^{2} \end{matrix}$
$\begin{matrix} \frac{a^{2} - 2 a b + b^{2}}{a^{2} - a b} p a r a b & = a \\ \frac{a^{2} - 2 a b + b^{2}}{a^{2} - a b} & = \frac{{(a - b)}^{2}}{a (a - b)} \\ = \frac{a - b}{a} \\ = \frac{a - a}{a} \\ = 0 \end{matrix}$
$\begin{matrix} \frac{x^{2} - y^{2}}{x y - y^{2}} y & = x \\ \frac{x^{2} - y^{2}}{x y - y^{2}} & = \frac{(x - y) (x + y)}{y (x - y)} \\ = \frac{x + y}{y} \\ = \frac{x + x}{x} \\ = \frac{2 x}{x} \\ = 2 \end{matrix}$
$\begin{matrix} \frac{x^{3} - a^{3}}{a^{2} x - a^{3}} p a r a x & = a \\ \frac{x^{3} - a^{3}}{a^{2} x - a^{3}} & = \frac{(x - a) (x^{2} + a x + a^{2})}{a^{2} (x - a)} \\ = \frac{x^{2} + a x + a^{2}}{a^{2}} \\ = \frac{a^{2} + a (a) + a^{2}}{a^{2}} \\ = \frac{3 a^{2}}{a^{2}} \\ = 3 \end{matrix}$
$\begin{matrix} \frac{8 x^{2} + 6 x - 9}{12 x^{2} - 13 x + 3} p a r a x & = \frac{3}{4} \\ \frac{8 x^{2} + 6 x - 9}{12 x^{2} - 13 x + 3} & = \frac{8 x^{2} + 12 x - 6 x - 9}{12 x^{2} - 4 x - 9 x + 3} \\ = \frac{4 x (2 x + 3) - 3 (2 x + 3)}{4 x (3 x - 1) - 3 (3 x - 1)} \\ = \frac{(4 x - 3) (2 x + 3)}{(4 x - 3) (3 x - 1)} \\ = \frac{2 x + 3}{3 x - 1} \\ = \frac{2 (\frac{3}{}) + 3}{3 (\frac{3}{4}) - 1} \\ = \frac{\frac{3}{2} + 3}{\frac{9}{4} - 1} \\ = \frac{\frac{3 + 6}{2}}{\frac{9 - 4}{}} \\ = \frac{9}{\frac{5}{2}} \\ = \frac{18}{5} \end{matrix}$
$\begin{matrix} \frac{x^{3} + 6 x^{2} + 12 x + 8}{x^{4} - 8 x^{2} + 16} p a r a x & = - 2 \\ \frac{x^{3} + 6 x^{2} + 12 x + 8}{x^{4} - 8 x^{2} + 16} & = \frac{(x^{3} + 8) + (6 x^{2} + 12 x)}{{(x^{2} - 4)}^{2}} \\ = \frac{(x + 2) (x^{2} - 2 x + 4) + 6 x (x + 2)}{{[(x - 2) (x + 2)]}^{2}} \\ = \frac{(x + 2) [(x^{2} - 2 x + 4) + 6 x]}{{(x - 2)}^{2} {(x + 2)}^{2}} \\ = \frac{x^{2} + 4 x + 4}{{(x - 2)}^{2} (x + 2)} \\ = \frac{{(x + 2)}^{2}}{{(x - 2)}^{2} (x + 2)} \\ = \frac{x + 2}{{(x - 2)}^{2}} \\ = \frac{- 2 + 2}{{(- 2 - 2)}^{2}} \\ = 0 \end{matrix}$
$\begin{matrix} \frac{9 x^{3} + 3 x^{2} + 3 x + 1}{27 x^{3} + 1} p a r a x & = - \frac{1}{3} \\ \frac{9 x^{3} + 3 x^{2} + 3 x + 1}{27 x^{3} + 1} & = \frac{3 x^{2} (3 x + 1) + (3 x + 1)}{(3 x + 1) (9 x^{2} - 3 x + 1)} \\ = \frac{(3 x^{2} + 1) (3 x + 1)}{(3 x + 1) (9 x^{2} - 3 x + 1)} \\ = \frac{3 x^{2} + 1}{9 x^{2} - 3 x + 1} \\ = \frac{3 {(- \frac{1}{3})}^{2} + 1}{9 {(- \frac{1}{3})}^{2} - 3 (- \frac{1}{3}) + 1} \\ = \frac{3 (\frac{1}{3^{2}}) + 1}{9 (\frac{1}{3^{2}}) + 1 + 1} \\ = \frac{\frac{1}{3} + 1}{3} \\ = \frac{\frac{1 + 3}{3}}{3} \\ = \frac{4}{9} \end{matrix}$
$\begin{matrix} \frac{1}{x - 1} - \frac{3}{x^{3} - 1} p a r a x & = 1 \\ \frac{1}{x - 1} - \frac{3}{x^{3} - 1} & = \frac{x^{2} + x + 1 - 3}{(x - 1) (x^{2} + x + 1)} \\ = \frac{x^{2} + x - 2}{(x - 1) (x^{2} + x + 1)} \\ = \frac{(x + 2) (x - 1)}{(x - 1) (x^{2} + x + 1)} \\ = \frac{x + 2}{x^{2} + x + 1} \\ = \frac{1 + 2}{1^{2} + 1 + 1} \\ = \frac{3}{3} \\ = 1 \end{matrix}$
$\begin{matrix} (x^{2} + 3 x - 10) (1 + \frac{1}{x - 2}) p a r a x & = 2 \\ (x^{2} + 3 x - 10) (1 + \frac{1}{x - 2}) & = (x + 5) (x - 2) (\frac{x - 2 + 1}{x - 2}) \\ = (x + 5) (x - 1) \\ = (2 + 5) (2 - 1) \\ = 7 (1) \\ = 7 \end{matrix}$