Valor absoluto: Ecuaciones - FIAR

Introducción
Aclaraciones
Ecuaciones
Inecuaciones
Funciones

01. $\displaystyle \text {Calcula el valor de la expresión } 2+|2x-3|-|x-1| \text { , en } x=2$

$\displaystyle \bf 2+|2x-3|-|x-1| \text { , en } x=2 \medskip$
$\displaystyle 2+|2 \cdot 2-3|-|2-1|=2+1-1=2$

02. $\displaystyle \text {Calcula el valor de la expresión } 2x-2-|2x-5| \text { , en } x=-3$

$\displaystyle \bf 2x-2-|2x-5| \text { , en } x=-3 \medskip$
$\displaystyle 2(-3)-2-|2(-3)-5|=-6-2-11=-19$

03. $\displaystyle \text {Calcula el valor de la expresión } \frac{2 x-3|3 x-1|+|2 x-3|}{2|x|-3|x-4|} \text { , en } x=-1$

$\displaystyle \bf \frac{2 x-3|3 x-1|+|2 x-3|}{2|x|-3|x-4|} \text { , en } x=-1 \medskip$
$\displaystyle \frac{2(-1)-3|3(-1)-1|+|2(-1)-3|}{2|-1|-3|-1-4|}=\frac{-2-3 \cdot 4+5}{2-3 \cdot 5}=\frac{-9}{-13}=\frac{9}{13}$

04. $\displaystyle \text {Desarrolla la expresión } |2x-4|+x \text { eliminando los valores absolutos}$

$\displaystyle \bf |2x-4|+x= \medskip$

(1) $\begin{equation*} \displaystyle =\left\{\begin{array}{l1} -2 x+4+x & \text { si } 2 x-4<0 \\ 2 x-4+x & \text { si } 2 x-4 \geq 0 \end{array}= \begin{cases}4-x & \text { si } x<2 \\ 3 x-4 & \text { si } x \geq 2\end{cases}\right. \end{equation*}$

05. $\displaystyle \text {Desarrolla la expresión } x+|2x| \text { eliminando los valores absolutos}$

$\displaystyle \bf x+|2x|= \medskip$

(2) $\begin{equation*} \displaystyle =\left\{\begin{array}{ll} x-2 x & \text { si } 2 x<0 \\ x+2 x & \text { si } 2 x \geq 0 \end{array}=\left\{\begin{array}{lll} -x & \text { si } & x<0 \\ 3 x & \text { si } & x \geq 0 \end{array}\right.\right. \end{equation*}$

06. $\displaystyle \text {Desarrolla la expresión } |x-1|+x \text { eliminando los valores absolutos}$

$\displaystyle \bf |x-1|+x= \medskip$

(3) $\begin{equation*} \displaystyle =\left\{\begin{array}{ll} -x+1+x & \text { si } x-1<0 \\ x-1+x & \text { si } x-1 \geq 0 \end{array}= \begin{cases}1 & \text { si } x<1 \\ 2 x-1 & \text { si } x \geq 1\end{cases}\right. \end{equation*}$

07. $\displaystyle \text {Desarrolla la expresión } (x-2)^2-|x-2| \text { eliminando los valores absolutos}$

$\displaystyle \bf (x-2)^2-|x-2|= \medskip$

(4) $\begin{equation*} \displaystyle =\left\{\begin{array}{ll} x^{2}-4 x+4+x-2 \\ x^{2}-4 x+4-x+2 \end{array}= \begin{cases}x^{2}-3 x+2 & \text { si } x<2 \\ x^{2}-5 x+6 & \text { si } x \geq 2\end{cases}\right. \end{equation*}$

08. $\displaystyle |2x-1|-x=2$

$\displaystyle \bf |2x-1|-x=2 \medskip$
$\displaystyle \Rightarrow \left\{\begin{array} { l l } { - 2 x + 1 - x = 2 } & { \text { si } 2 x - 1 < 0 } \\ { 2 x - 1 - x = 2 } & { \text { si } 2 x - 1 \geq 0 } \end{array}\right. \medskip$
$\displaystyle \Rightarrow \left\{\begin{array}{ll} x=-\frac{1}{3} & \text { si } x<\frac{1}{2} \\ x=3 & \text { si } x \geq \frac{1}{2} \end{array} \right. \medskip$
$\displaystyle \Rightarrow x=-\frac{1}{3}, x=3$

09. $\displaystyle |x+6|=2x+6$

$\displaystyle \bf |x+6|=2x+6 \medskip$
$\displaystyle \Rightarrow \left\{\begin{array} { l l l } { - 3 x + 1 - 2 x = 1 1 } & { \text { si } } & { 3 x - 1 < 0 } \\ { 3 x - 1 - 2 x = 1 1 } & { \text { si } } & { 3 x - 1 \geq 0 } \end{array}\right. \medskip$
$\displaystyle \Rightarrow \left\{\begin{array}{ll} \displaystyle x=-2 & \text { si } x<\frac{1}{3} \\ \displaystyle x=12 & \text { si } x \geq \frac{1}{3} \end{array} \Rightarrow x=-2, x=12\right.$

10. $\displaystyle |x-2|+|x-3|=9$

$\displaystyle \bf |x-2|+|x-3|=9 \medskip$
$\displaystyle \Rightarrow\left\{\begin{array} { l l } { - x + 2 - x + 3 = 9 } & { \text { si } x < 2 } \\ { x - 2 - x + 3 = 9 } & { \text { si } 2 \leq x < 3 } \\ { x - 2 + x - 3 = 9 } & { \text { si } x \geq 3 } \end{array} \right. \medskip$
$\displaystyle \Rightarrow \left\{\begin{array}{ll} x=-2 & \text { si } x<2 \\ 0=8 & \text { si } 2 \leq x<3 \Rightarrow x=-2, x=7 \\ x=7 & \text { si } x \geq 3 \end{array}\right.$

11. $\displaystyle 3|x-1|+5|x-1|-2|x-1|=6$

12. $\displaystyle |x-1|+|x+1|-1=|x|$

13. $\displaystyle \left|2-\frac{x}{3}\right|=2}$

Solución pendiente

14. $\displaystyle \left|x-\frac{1}{2}\right|+2x=\frac{1}{2}$

$\displaystyle \bf \left|x-\frac{1}{2}\right|+2x=\frac{1}{2} \medskip$
$\Rightarrow \left\{\begin{array} { l l } { - x + \frac { 1 } { 2 } + 2 x = \frac { 1 } { 2 } } & { \text { si } x - \frac { 1 } { 2 } < 0 } \\ { x - \frac { 1 } { 2 } + 2 x = \frac { 1 } { 2 } } & { \text { si } x - \frac { 1 } { 2 } \geq 0 } \end{array}\right. \medskip$
$\Rightarrow \left\{\begin{array}{ll} x=0 & \text { si } x<\frac{1}{2} \\ x=\frac{1}{3} & \text { si } x \geq \frac{1}{2} \end{array} \Rightarrow x=0\right.$

15. $\displaystyle \left|\frac{4-x}{3x}\right|=3$

Solución pendiente

16. $\displaystyle \left|\frac{x+1}{x-5}\right|=1$

Solución pendiente

17. $\displaystyle \left|\frac{2x-3}{1-x}\right|=2$

Solución pendiente

Deja un comentarioCancelar respuesta