Ejercicio 252

octubre 30, 2019abril 14, 2019 por Gestor Baldor

Comparte esto 👍👍

CAPITULO XXXI

R a d i c a l e s
Resolución de Ecuaciones con radicales que se reducen a primer grado

Ejercicio252

Resolver las ecuaciones:

$\begin{matrix} \sqrt{x} + \sqrt{x + 5} & = \frac{10}{\sqrt{x}} \\ \sqrt{x^{2}} + \sqrt{x (x + 5)} & = 10 \\ x + \sqrt{x (x + 5)} & = 10 \\ {(\sqrt{x (x + 5)})}^{2} & = {(10 - x)}^{2} \\ x^{2} + 5 x & = 100 - 20 x + x^{2} \\ 5 x + 20 x & = 100 \\ 25 x & = 100 \\ x & = \frac{}{25} \\ x & = 4 \end{matrix}$
$\begin{matrix} \sqrt{4 x - 11} + 2 \sqrt{x} & = \frac{55}{\sqrt{4 x - 11}} \\ \sqrt{{(4 x - 11)}^{2}} + 2 \sqrt{x (4 x - 11)} & = 55 \\ 4 x - 11 + 2 \sqrt{x (4 x - 11)} & = 55 \\ 2 \sqrt{x (4 x - 11)} & = 66 - 4 x \\ 2 \sqrt{x (4 x - 11)} & = 2 (33 - 2 x) \\ {(\sqrt{x (4 x - 11)})}^{2} & = {(33 - 2 x)}^{2} \\ 4 x^{2} - 11 x & = 1089 - 132 x + 4 x^{2} \\ 132 x - 11 x & = 1089 \\ 121 x & = 1089 \\ x & = \frac{}{121} \\ x & = 9 \end{matrix}$
$\begin{matrix} \sqrt{x} - \sqrt{x - 7} & = \frac{4}{\sqrt{x}} \\ \sqrt{x^{2}} - \sqrt{x (x - 7)} & = 4 \\ x - \sqrt{x^{2} - 7 x} & = 4 \\ {(x - 4)}^{2} & = {(\sqrt{x^{2} - 7 x})}^{2} \\ x^{2} - 8 x + 16 & = x^{2} - 7 x \\ - 8 x + 7 x & = - 16 \\ - x & = - 16 \\ x & = 16 \end{matrix}$
$\begin{matrix} \frac{\sqrt{x} - 2}{\sqrt{x} + 4} & = \frac{\sqrt{x} + 1}{\sqrt{x} + 13} \\ (\sqrt{x} - 2) (\sqrt{x} + 13) & = (\sqrt{x} + 1) (\sqrt{x} + 4) \\ \sqrt{x^{2}} + 13 \sqrt{x} - 2 \sqrt{x} - 26 & = \sqrt{x^{2}} + 4 \sqrt{x} + \sqrt{x} + 4 \\ 11 \sqrt{x} - 5 \sqrt{x} & = 26 + 4 \\ 6 \sqrt{x} & = 30 \\ \sqrt{x} & = \frac{}{6} \\ {(\sqrt{x})}^{2} & = 5^{2} \\ x & = 25 \end{matrix}$
$\begin{matrix} \frac{6}{\sqrt{x + 8}} & = \sqrt{x + 8} - \sqrt{x} \\ 6 & = {(\sqrt{x + 8})}^{2} - \sqrt{x (x + 8)} \\ 6 & = x + 8 - \sqrt{x^{2} + 8 x} \\ \sqrt{x^{2} + 8 x} & = x + 8 - 6 \\ {(\sqrt{x^{2} + 8 x})}^{2} & = {(x + 2)}^{2} \\ x^{2} + 8 x & = x^{2} + 4 x + 4 \\ 8 x - 4 x & = 4 \\ 4 x & = 4 \\ x & = \frac{4}{4} \\ x & = 1 \end{matrix}$
$\begin{matrix} \sqrt{x - 3} + \frac{8}{\sqrt{x + 9}} & = \sqrt{x + 9} \\ \frac{\sqrt{(x - 3) (x + 9)} + 8}{\sqrt{x + 9}} & = \sqrt{x + 9} \\ \sqrt{x^{2} + 6 x - 27} + 8 & = \sqrt{{(x + 9)}^{2}} \\ \sqrt{x^{2} + 6 x - 27} + 8 & = x + 9 \\ {(\sqrt{x^{2} + 6 x - 27})}^{2} & = {(x + 1)}^{2} \\ x^{2} + 6 x - 27 & = x^{2} + 2 x + 1 \\ 6 x - 2 x & = 27 + 1 \\ 4 x & = 28 \\ x & = \frac{}{4} \\ x & = 7 \end{matrix}$
$\begin{matrix} \frac{\sqrt{x} + 4}{\sqrt{x} - 2} & = \frac{\sqrt{x} + 11}{\sqrt{x} - 1} \\ (\sqrt{x} + 4) (\sqrt{x} - 1) & = (\sqrt{x} + 11) (\sqrt{x} - 2) \\ \sqrt{x^{2}} - \sqrt{x} + 4 \sqrt{x} - 4 & = \sqrt{x^{2}} - 2 \sqrt{x} + 11 \sqrt{x} - 22 \\ 3 \sqrt{x} & = 9 \sqrt{x} - 22 + 4 \\ - 6 \sqrt{x} & = - 18 \\ \sqrt{x} & = \frac{-}{- 6} \\ {(\sqrt{x})}^{2} & = 3^{2} \\ x & = 9 \end{matrix}$
$\begin{matrix} 2 \sqrt{x + 6} - \sqrt{4 x - 3} & = \frac{9}{\sqrt{4 x - 3}} \\ 2 \sqrt{(x + 6) (4 x - 3)} - \sqrt{{(4 x - 3)}^{2}} & = 9 \\ 2 \sqrt{4 x^{2} - 3 x + 24 x - 18} - 4 x + 3 & = 9 \\ 2 \sqrt{4 x^{2} + 21 x - 18} & = 9 - 3 + 4 x \\ 2 {(\sqrt{4 x^{2} + 21 x - 18})}^{2} & = 2 {(2 x + 3)}^{2} \\ 4 x^{2} + 21 x - 18 & = 4 x^{2} + 12 x + 9 \\ 21 x - 12 x & = 18 + 9 \\ 9 x & = 27 \\ x & = \frac{}{9} \\ x & = 3 \end{matrix}$
$\begin{matrix} \frac{\sqrt{x} - 2}{\sqrt{x} + 2} & = \frac{2 \sqrt{x} - 5}{2 \sqrt{x} - 1} \\ (\sqrt{x} - 2) (2 \sqrt{x} - 1) & = (2 \sqrt{x} - 5) (\sqrt{x} + 2) \\ 2 \sqrt{x^{2}} - \sqrt{x} - 4 \sqrt{x} + 2 & = 2 \sqrt{x^{2}} + 4 \sqrt{x} - 5 \sqrt{x} - 10 \\ 2 + 10 & = 4 \sqrt{x} \\ 12 & = 4 \sqrt{x} \\ {(\frac{}{4})}^{2} & = {(\sqrt{x})}^{2} \\ x & = 9 \end{matrix}$
$\begin{matrix} \sqrt{x + 14} - \sqrt{x - 7} & = \frac{6}{\sqrt{x - 7}} \\ \sqrt{(x + 14) (x - 7)} - \sqrt{{(x - 7)}^{2}} & = 6 \\ \sqrt{x^{2} + 7 x - 98} - x + 7 & = 6 \\ \sqrt{x^{2} + 7 x - 98} & = 6 - 7 + x \\ {(\sqrt{x^{2} + 7 x - 98})}^{2} & = {(x - 1)}^{2} \\ x^{2} + 7 x - 98 & = x^{2} - 2 x + 1 \\ 7 x + 2 x & = 98 + 1 \\ 9 x & = 99 \\ x & = \frac{}{9} \\ x & = 11 \end{matrix}$