Skip to main content
Global

5.4: 零指数规则

  • Page ID
    171053
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    在第 5.3 节中,分子中数字的指数始终大于分母中数字的指数。

    在第 5.4 节中,分子中数字的指数将等于分母中数字的指数。

    定义:零指数规则

    对于任何实数\(a\),零指数规则如下

    \(a^0= 1\)

    想法:

    来自前面的章节:

    \[x^5 = x \cdot x \cdot x \cdot x \cdot x \nonumber \]

    \[\dfrac{x^5 }{x^5} =\dfrac{ x \cdot x \cdot x \cdot x \cdot x }{x \cdot x \cdot x \cdot x \cdot x }= \dfrac{\cancel{x \cdot x \cdot x \cdot x \cdot x }}{\cancel{x \cdot x \cdot x \cdot x \cdot x }}= 1 \nonumber \]

    因此,

    \[\dfrac{x ^5 }{x^5} = x^{5−5 }= x^0=1 \nonumber \]

    使用零指数规则来简化表达式。

    1. \(\dfrac{x^9 }{x^9}\)
    2. \(\dfrac{d^5 }{d^2 \cdot d^3}\)
    3. \(\dfrac{5(xy)^3 }{(xy)^3}\)
    4. \(-\dfrac{y^3 }{\sqrt{5}y^3}\)
    5. \(\dfrac{(ab^2 )^7 }{(ab^2)^2 \cdot (ab^2)^4 ˙(ab^2)}\)
    解决方案
    表情 零指数规则
    \(\dfrac{x^9 }{x^9}\) \(x^{9−9} = x^0 = 1\)
    \(\dfrac{d^5 }{d^2 \cdot d^3}\) \(\dfrac{d^5 }{d^{2+3 }}= \dfrac{d^5 }{d^5} = d^{5−5 }= d^{0} = 1\)
    \(\dfrac{5(xy)^3 }{(xy)^3}\)

    \(5 \cdot \dfrac{(xy)^3 }{(xy)^3 }= 5 \cdot (xy)^{3−3 }= 5 \cdot (xy)^0 = 5 \cdot 1 = 5 \)

    可以分解常数 5,以便清楚地看到共同的基础。

    \(-\dfrac{y^3 }{\sqrt{5}y^3}\)

    \(− \dfrac{1}{ \sqrt{5}} \cdot \dfrac{y^3 }{y^3 }= − \dfrac{1}{ \sqrt{5}} \cdot y ^{3−3 }= − \dfrac{1}{ \sqrt{5}} \cdot y^0 = − \dfrac{1}{ \sqrt{5}} \cdot 1 = − \dfrac{1}{ \sqrt{5}}\)

    可以将常量分解出来\(−\left( \dfrac{1 }{\sqrt{5}}\right )\),以便清楚地看到共同的基础。

    \(\dfrac{(ab^2 )^7 }{(ab^2)^2 \cdot (ab^2)^4 ˙(ab^2)}\)

    \(\dfrac{(ab^2 )^7 }{(ab^2)^{2+4+1}}= \dfrac{(ab^2 )^7}{ (ab^2)^7} = (ab^2 )^{7−7 }= (ab^2 )^0 = 1\)

    首先,使用指数乘积法则简化分母。 然后使用指数的商法则来简化其余表达式。

    注意:\(0^0\)不等于 1。 这是高级课程中涵盖的特例。 现在\(0^0\)考虑未定义。

    使用指数简化表达式的有用步骤

    1. 确定共同基础。
    2. 如果需要,使用指数乘积法则组合常用基数。
    3. 如果表达式在分子和分母中都包含公基,则根据需要使用指数的商法则。

    使用本章到目前为止涵盖的所有指数规则来简化以下内容。

    1. \(\dfrac{z ^4 }{z^ 4}\)
    2. \(\dfrac{d^2 \cdot d^8}{ d^7 \cdot d^3}\)
    3. \(\dfrac{5(x + y)^3 }{2(x + y)^3}\)
    4. \(−\dfrac{\sqrt{9}{y^3 }}{y^3}\)
    5. \(\dfrac{(a^3b^2 )^9}{ (a^3b^2)^3 \cdot (a^3b^2)^4 ˙(a^3b^2)^2}\)
    6. \(\dfrac{(xyz)^{19} }{(xyz)^{19}}\)
    Template:AutoNumHeadings

    This page titled 5.4: 零指数规则 is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Victoria Dominguez, Cristian Martinez, & Sanaa Saykali (ASCCC Open Educational Resources Initiative) .