9.1: Simplificar expresiones racionales
- Page ID
- 112594
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Una expresión racional se escribe como cociente de polinomios
\[\dfrac{P(x)}{Q(x)} \nonumber \]
donde\(P(x)\) y\(Q(x)\) son polinomios en una variable\(x\).
Para simplificar una expresión racional, factorizar tanto el numerador como el denominador, y eliminar los factores comunes tanto del numerador como del denominador. Una expresión racional simplificada tiene una sola división, y un solo numerador y denominador. Si las expresiones no pueden ser factorizadas, entonces la expresión racional no puede simplificarse.
Simplifica las expresiones racionales:
- \(\dfrac{x^2 + 2x − 3}{x^2 + 4x + 3}\)
- \(\dfrac{(x^2 + 1)^2 (−2) + (2x)(2)(x^2 + 1)(2x)}{(x^2 + 1)^4}\)
- \(\dfrac{(x^2 + 1) \frac{1}{2} (x^{−\frac{1}{2}}) − (2x)(x^{\frac{1}{2}})}{(x^2 + 1)^2}\)
Solución
- \(\begin{array} &&\dfrac{x^2 + 2x − 3}{x^2 + 4x + 3} &\text{Example problem} \\ &\dfrac{(x + 3)(x − 1)}{(x + 3)(x + 1)} &\text{Factor both numerator and denominator.} \\ &\dfrac{\cancel{(x + 3)}(x − 1)}{\cancel{(x + 3)}(x + 1)} &\text{Remove common factors, because \(\dfrac{x + 3}{x + 3} = 1\)}\\ &\ dfrac {x − 1} {x + 1} &\ text {Respuesta final}\ end {array}\)
- \(\begin{array} &&\dfrac{(x^2 + 1)^2 (−2) + (2x)(2)(x^2 + 1)(2x)}{(x^2 + 1)^4} &\text{Example problem} \\ &\dfrac{2(x^2 + 1)[(x^2 + 1)(−1) + (2x)(2x)]}{(x^2 + 1)^4} &\text{Factor out 2(x^2 + 1)} \\ &\dfrac{2 \cancel{(x^2 + 1)}[(x^2 + 1)(−1) + (2x)(2x)]}{\cancel{(x^2+1)}(x^2 + 1)^3} &\text{Remove common factors, because \(\dfrac{x^2 + 1}{x^2 + 1} = 1\)}\\ &\ dfrac {2 [−x^2 − 1 + 4x^2]} {(x^2 + 1) ^3} &\ text {Simplifica multiplicando y combinando términos similares}\\ &\ dfrac {2 (3x^2 − 1)} {(x^2 + 1) ^3} &\ text {Respuesta final}\ end {array}\)
- \(\begin{array} &&\dfrac{(x^2 + 1) \frac{1}{2} (x^{−\frac{1}{2}}) − (2x)(x^{\frac{1}{2}})}{(x^2 + 1)^2} &\text{Example problem} \\ &\dfrac{\frac{(x^2+1)}{2x^{\frac{1}{2}}} − (2x)(x^{ \frac{1}{2} })}{(x^2 + 1)^2} &\text{Work with the negative exponent in the first term of the numerator by moving the factor to the denominator of the first term, next to the \(2\).}\\ &\ dfrac {(x^2 + 1) − (2x) (x^ {\ frac {1} {2}}) 2 (x^ {\ frac {1} {2}})} {\ dfrac {2x^ {\ frac {1} {2}}} {(x^2 + 1) ^2}} &\ text {denominador común}\ &\ dfrac {x^2 + 1 − 4x^2} {(2x^ {\ frac {1} {2}}) (x^2 + 1) ^2} &\ text {Simplifica multiplicando y combinando términos similares}\\ &\ dfrac {−3x^2 + 1} {(2x^ {\ frac {1} {2}}) (x^2 + 1) ^2} &\ text {Respuesta final}\ end {array}\)
Simplifica las expresiones racionales:
- \(\dfrac{2x^2 + 3x − 2}{2x^2 + 5x − 3}\)
- \(\dfrac{(t^2 + 4)(2t − 4) − (t^2 − 4t + 4)(2t)}{(t^2 + 4)^2}\)
- \(\dfrac{(2)(x − 4)(x^2 + 4x + 4)}{(x + 2)(x^2 − 16)}\)
- \(\dfrac{12x^2 + 19x − 21}{12x^2 + 38x − 40}\)