1.2: El valor absoluto
- Page ID
- 117751
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)El valor absoluto de un número real\(c\), denotado por\(|c|\) el número no negativo que es igual en magnitud (o tamaño) a\(c\), es decir, es el número resultante de ignorar el signo:
\ [|c|=\ left\ {\ begin {array} {cl}
c, &\ text {if} c\ text {es positivo o cero}\\
-c, &\ text {if} c\ text {es negativo}
\ end {array}\ right. \ nonumber\]
\(|-4|=4\)
\(|12|=12\)
\(|-3.523|=3.523\)
¿Para\(x\) qué números reales tienes\(|x|=3\)?
Solución
Desde\(|3|=3\) y\(|-3|=3\), vemos que hay dos soluciones,\(x=3\) o\(x=-3\).
El conjunto de soluciones es\(S=\{-3,3\}\).
|x|=5] Resolver para\(x\):\(|x|=5\)
Solución
\(x=5\)o\(x=-5\). El conjunto de soluciones es\(S=\{-5,5\}\).
|x|=-7] Resolver para\(x\):\(|x|=-7\).
Solución
Tenga en cuenta\(|7|=7\) eso\(|-7|=7\) y para que estos no puedan dar ninguna solución. En efecto, no hay soluciones, ya que el valor absoluto es siempre no negativo. El conjunto de soluciones es el conjunto vacío\(S=\{\}\).
Resolver para\(x\):\(|x|=0\).
Solución
Ya que\(-0=0\), sólo hay una solución,\(x=0\). Por lo tanto,\(S=\{0\}\).
Resolver para\(x\):\(|x+2|=6\).
Solución
Ya que el valor absoluto de\(x+2\) es\(6\), vemos que\(x+2\) tiene que ser cualquiera\(6\) o\(-6\).
Evaluamos cada caso,
\ [\ begin {array} {l|l}
\ text {ya sea} x+2=6, &\ text {o} x+2=-6\\
\ Longrightarrow x=6-2, &\ LongRightarrow x=-6-2\
\ LongRightarrow x=4, &\ LongRightarrow x=-8
\ end {array}\ nonumber\]
El conjunto de soluciones es\(S=\{-8,4\}\).
Resolver para\(x\):\(|3x-4|=5\)
Solución
\ [\ begin {array} {l|l}
\ text {O} 3 x-4=5 &\ text {o} 3 x-4=-5\\
\ Longrightarrow 3 x=9 &\ Longrightarrow 3 x=-1\
\ LongRightarrow x=3 &\ Rightarrow x=-\ frac {1} {3}
\ end {array}\ nonumber\]
El conjunto de soluciones es\(S=\{-\frac 1 3,3\}\).
Resolver para\(x\):\(-2\cdot |12+3x|=-18\)
Solución
Dividiendo ambos lados por\(-2\) da\(|12+3x|=9\). Con esto, tenemos los dos casos
\ [\ begin {array} {l|l}
\ text {O} 12+3 x=9 &\ text {o} 12+3 x=-9\\
\ Longrightarrow 3 x=-3 &\ Longrightarrow 3 x=-21\
\ LongRightarrow x=-1 &\ LongRightarrow x=-7
\ end {array}\ nonumber\]
El conjunto de soluciones es\(S=\{-7,-1\}\).