9.4: 合理化代数分数
- Page ID
- 171281
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)如果有理表达式的分母包含涉及激进的总和或差，则最好始终通过将分子和分母乘以分母的共轭来合理化分母。
分母的共轭包含相同的项，但运算相反（加法或减法）。
合理化分母并简化：
- \(\dfrac{1}{1 − \sqrt{x}}\)
- \(\dfrac{1}{\sqrt{x} − \sqrt{y}}\)
- \(\dfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} − \sqrt{y}}\)
解决方案
- \(\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 分母。}\\ &\ dfrac {1 +\ sqrt {x}} {1 −\ 取消 {\ sqrt {x}} − (\ sqrt {x}) ^2} &\ text {移除总和为零的相反项。}\\ &\ dfrac {1 +\ sqrt {x}} {1 − x} &\ text {方块的根\(x\)，数量平方为\(x\)。}\\ &\ dfrac {1 +\ sqrt {x}} {1 − x} &\ text {分母合理化后的最终答案，这意味着分母中没有平方根项。} \ 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 {y} − (\ sqrt {y}) ^2} &\ text {FOIL 分母。}\\ &\ dfrac {(\ sqrt {x} +\ sqrt {y})} {(\ sqrt {x}) ^2 −\ 取消 {\ sqrt {x}\ sqrt {y}} +\ cancel {\ sqrt {y}} − (\ sqrt {y}) ^2} &\ text {删除相反的术语总和为零。}\\ &\ dfrac {\ sqrt {x} +\ sqrt {y}} {x − y} &\ text {数量的平方根为\(x\)，数量平方为\(y\)，平方根为\(y\)。}\\ &\ dfrac {\ sqrt {x} +\ sqrt {y} {x} {x − y}} {x − y} &\ text {合理化了，意思是\(x\)分母中没有平方根项。} \ 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 {y} − (\ sqrt {y}) ^2} &\ text {FOIL 分子和分母。}\\ &\ dfrac {x + 2\ sqrt {x}\ sqrt {y}\ sqrt {y} {(\ sqrt {x}) ^2 −\ 取消 {\ sqrt {x}\ sqrt {y}} − (\sqrt {y}) ^2} &\ text {移除总和为零的相反项。}\\ &\ dfrac {x + 2\ sqrt {x}\ sqrt {y} {x − y} {x − y} &\ text {x}\ sqrt {y} + y} {x − y} &\\(x\)\(x\)\(y\)\(y\)text {对分母进行了合理化的最终答案，这意味着分母中没有平方根项。} \ end {array}\)
合理化分母并简化：
- \(\dfrac{x}{1 − \sqrt{x}}\)
- \(\dfrac{1}{1 − \sqrt{x}}\)
- \(\dfrac{2 \sqrt{x}}{\sqrt{x} − 1}\)
- \(\dfrac{x − 1}{\sqrt{x} − 1}\)