9.4: Racionalizar fracciones algebraicas
- Page ID
- 112582
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Si el denominador de una expresión racional contiene sumas o diferencias que involucran radicales, es buena forma racionalizar siempre el denominador multiplicando el numerador y denominador por el conjugado del denominador.
El conjugado del denominador contiene los mismos términos, pero operaciones opuestas (suma o resta).
Racionalizar el denominador y simplificar:
- \(\dfrac{1}{1 − \sqrt{x}}\)
- \(\dfrac{1}{\sqrt{x} − \sqrt{y}}\)
- \(\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} − \sqrt{y}}\)
Solución
- \(\begin{array} &&\dfrac{1}{1 − \sqrt{x}} &\text{Example problem} \\ &\dfrac{(1)(1 + \sqrt{x})}{(1 − \sqrt{x})(1 + \sqrt{x})} &\text{Multiply both numerator and denominator by the conjugate, which is \((1+\sqrt{x})\)}\\ &\ dfrac {1 +\ sqrt {x}} {1 −\ sqrt {x} +\ sqrt {x} − (\ sqrt {x}) ^2} &\ text {FOIL el denominador.}\\ &\ dfrac {1 +\ sqrt {x}} {1 −\ cancel {\ sqrt {x}} +\ cancel {\ sqrt {x}}} − (\ sqrt {x}) ^2} &\ text {Eliminar términos opuestos que suman a cero.}\\ &\ dfrac {1 +\ sqrt {x}} {1 − x} &\ text {El cuadrado raíz de\(x\), cantidad al cuadrado es\(x\).}\\ &\ dfrac {1 +\ sqrt {x}} {1 − x} &\ text {Respuesta final con el denominador racionalizado, es decir, que no hay términos de raíz cuadrada en el denominador.} \ end {array}\)
- \(\begin{array} &&\dfrac{1}{\sqrt{x} − \sqrt{y}} &\text{Example problem} \\ &\dfrac{(1)(\sqrt{x} + \sqrt{y})}{(\sqrt{x} − \sqrt{y})(\sqrt{x} + \sqrt{y})} &\text{Multiply both numerator and denominator by the conjugate, which is \((\sqrt{x} + \sqrt{y})\)}\\ &\ dfrac {(\ sqrt {x} +\ sqrt {y})} {(\ sqrt {x}) ^2 −\ sqrt {x}\ sqrt {y} +\ sqrt {x}\ sqrt {y} − (\ sqrt {y}) ^2} &\ text {FOIL el denominador.}\\ &\ dfrac {(\ sqrt {y}) ^2} &\ text {FOIL el denominador.}\\ &\ dfrac {(\ sqrt {x} +\ sqrt {y})} {(\ sqrt {x}) ^2 −\ cancel {\ sqrt {x}\ sqrt {y}} +\ cancel {\ sqrt {x}\ sqrt {y}} − (\ sqrt {y}) ^2} &\ text {Eliminar términos opuestos que suma a cero.}\\ &\ dfrac {\ sqrt {x} +\ sqrt {y}} {x − y} &\ text {La raíz cuadrada de\(x\), cantidad cuadrada es\(x\), y la raíz cuadrada de\(y\), cantidad cuadrada es\(y\).}\\ &\ dfrac {\ sqrt {x} +\ sqrt {y}} {x y −} &\ text {Respuesta final con el denominador racionalizado, significado que no hay términos de raíz cuadrada en el denominador.} \ end {array}\)
- \(\begin{array} &&\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} − \sqrt{y}} &\text{Example problem} \\ &\dfrac{(\sqrt{x} + \sqrt{y})(\sqrt{x} + \sqrt{y})}{(\sqrt{x} − \sqrt{y})(\sqrt{x} + \sqrt{y})} &\text{Multiply both numerator and denominator by the conjugate, which is \((\sqrt{x} + \sqrt{y})\)}\\ &\ dfrac {(\ sqrt {x}) ^2 + 2 (\ sqrt {x}\ sqrt {y}) + (\ sqrt {y}) ^2} {(\ sqrt {x}) ^2 −\ sqrt {x}\ sqrt {y} +\ sqrt {x}\ sqrt {y} − (\ ^sqrt {y}) 2} &\ text {FOIL el numerador y el denominador.}\\ &\ dfrac {x + 2\ sqrt {x}\ sqrt {y} + y} {(\ sqrt {x}) ^2 −\ cancel {\ sqrt {x}\ sqrt {y}} +\ cancel {\ sqrt {x}\ sqrt {y}} − (\ sqrt {y}) ^2} &\ text {Eliminar términos opuestos que suman a cero.}\\ &\ dfrac {x + 2\ sqrt {x}\ sqrt {y} + y} {x − y} &\ text {La raíz cuadrada de\(x\), cantidad cuadrada es\(x\), y la raíz cuadrada de\(y\), cantidad cuadrada es\(y\).}\\ &\ dfrac {x + 2\ sqrt {x}\ sqrt {y} + y} {x − y} &\ text {Respuesta final con el denominador racionalizado, es decir, que no hay términos de raíz cuadrada en el denominador.} \ end {array}\)
Racionalizar el denominador y simplificar:
- \(\dfrac{x}{1 − \sqrt{x}}\)
- \(\dfrac{1}{1 − \sqrt{x}}\)
- \(\dfrac{2 \sqrt{x}}{\sqrt{x} − 1}\)
- \(\dfrac{x − 1}{\sqrt{x} − 1}\)