5.1 重新审视指数表达式-interactive

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• 选择节重审指数表达式

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  重审指数表达式

  Lesson Objectives

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  • Apply the  various  product and quotient  properties to write .
    ::应用各种产品和商数属性写入 。
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  • Apply the zero exponent property to write equivalent expressions.
    ::应用零引言属性来写入等效表达式。
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  • Apply the negative exponent property to write equivalent expressions.
    ::应用负表情属性写入等同表达式。
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  • Apply multiple properties of exponents to write equivalent expressions.  
    ::应用引言的多个属性来写入等同表达式 。

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  Introduction: Nanotechnology
  ::导言:纳米技术

  

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  Throughout history, humans have expanded their focus beyond Earth to the solar system, galaxy, and even universe. This focus on universal order has shaped mathematics. However, the focus outward isn't the only thing shaping the use of mathematics because  humans are  also focusing inward. Nanotechnology is used in many aspects of everyday life that you may not realize, like life-saving medicines, computer designs, and even the clothes you wear. The images above display a DNA lattice created by scientists on the left and a 3D printed part on the right. Negative exponents are commonly used to describe nano-sized objects of varying sizes. S cientists and engineers need to be able to perform all the same mathematical operations as they would on regular-sized objects, only with exponents. As a result, u nderstanding and applying the laws of exponents  can make it much easier to perform operations on two operands with exponents quickly .
  ::在整个历史中,人类的注意力已经从地球扩大到太阳系、银河系甚至宇宙。这种对宇宙秩序的注意力已经形成了数学。然而,外向的焦点并不是影响数学使用的唯一东西,因为人类也正在关注内向。纳米技术被用于日常生活的许多方面,而你可能没有意识到这一点,比如救生药物、计算机设计,甚至你穿的衣服。上面的图像显示的是科学家在左边创造的DNA阵列和右边打印的3D部分。消极的表征通常被用来描述不同尺寸的纳米大小的物体。科学家和工程师需要能够像在常规大小的物体上那样进行同样的数学操作,只有靠专家才能这样做。因此,理解和应用外来者的定律可以使在两个有前导力的歌剧上进行操作更容易得多。

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  Check out this interactive below to explore how  one of the basic laws is derived. 
  ::下面请查看下方的交互性, 以探讨其中一部基本法是如何产生的。

  

  INTERACTIVE

  Powers with the Same Term

  

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  • Move the points on the slider to observe taking the exponent of the expression.
    ::移动滑块上的点以观察表达式的缩略语。

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  Activity 1: Exponent Rules Review
  ::活动1:《指数规则审查》

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  Use the interactive below to review the basic exponent rules.
  ::使用以下互动方式审查基本提示规则。

   

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  Extension: Integer Powers
  ::扩展:整数强国

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  The  following interactive illustrates how the powers for like terms in the numerator and denominator of fractions with rational expressions are related and how they impact the resulting expression   when simplifying.
  ::以下互动说明带有合理表达的分数的分子和分母中类似术语的功率如何相关,以及在简化时如何影响所产生的表达方式。

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  This interactive will demonstrate how to simplify expressions with negative exponents and provide additional practice with this skill.
  ::这一互动将展示如何简化带有负面推手的表达方式,并提供这种技能的额外实践。

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  The following interactive provides you with an opportunity to explore the patterns among integer exponents . 
  ::以下互动为您提供了一个探索整数指数之间模式的机会。

    

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  Activity 2: Exponential Expressions
  ::活动2:指数表达式

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  Exponential expressions may require the combination of exponent rules to write an expression in its simplest form. As long as the exponent rules are followed correctly, there is not a set order for applying the rules.
  ::指数表达式可能需要将提示性规则组合在一起,才能以最简单的形式写出表达式。只要正确遵循提示性规则,则没有适用规则的固定顺序。

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  Example 
  ::示例示例示例示例

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  Simplify  $\begin{array}{r}(4x^{3}y)(3x^5{y}^2{)}^4\end{array}$
  ::简化 (4x3y (3x5y2) 4

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  To simplify this expression, you will need the following exponent rules:
  ::为简化这一表达式,您需要下列引文规则:

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    •   $\begin{array}{r}(ab{)}^n=(a^n)\cdot (b^n)\end{array}$
  ::• (ab)n=(an)_(bn)

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    •   %7D"> $\begin{array}{r}{\left(a^m\right)}^n=a^{(m)\cdot (n)}\end{array}$
  ::• (am)n=a(m)__

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    •   $\begin{array}{r}{a}^m\times a^n=a^{m+n}\end{array}$
  ::• AMXan=am+n

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  First, distribute the exponent to each factor of the term inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses . A ny factor with an exponent will require raising the power to a power. 
  ::首先,在括号内将引号分配到词句的每个因数。任何因数如果带有引号,都要求将权力赋予一个权力。

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  $$\begin{array}{rl}(4x^{3}y)(3x^5{y}^2{)}^4& =(4x^{3}y)(81x^{20}{y}^8)\end{array}$$

  
  :4x3y)(3x5y2)4=(4x3y)(81x20y8)

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  Then multiply like parts of each term by adding the exponents and multiplying the constants.
  ::然后,将各词各部分相仿地乘以累加功过簿和常数。

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  $$\begin{array}{r}(4x^{3}y)(81x^{20}{y}^8)=(4\cdot 81)x^{(3+20)}{y}^{(1+8)}\end{array}$$

  
  :4x3y)(81x20y8) = (481)x(3+20)y(1+8)

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  From here, simplify to obtain the answer.
  ::从这里,简化 以获得答案。

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  Answer:  $\begin{array}{r}324x^{23}{y}^9\end{array}$  
  ::答复:324x23y9

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  Example
  ::示例示例示例示例

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  Simplify the expression  $\begin{array}{r}{\left(\frac{x^{\text{-}2}y}{x^4{y}^3}\right)}^{\text{-}2}\end{array}$
  ::简化表达式 (x-2yx4y3)-2

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  To simplify this expression, you will need the following exponent rules:
  ::为简化这一表达式,您需要下列引文规则:

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    •   $\begin{array}{r}\frac{a^m}{a^n}=a^{m-n}\end{array}$
  ::• 男子=am-n

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    •   $\begin{array}{r}(ab{)}^n=(a^n)\cdot (b^n)\end{array}$
  ::• (ab)n=(an)_(bn)

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    •   $\begin{array}{r}{a}^{-m}=\frac{1}{a^m}\end{array}$
  ::• 上午1时

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  B egin by simplifying the expression in parenthesis.  First, use the quotient of powers exponent property to combine the $\begin{array}{r}x\end{array}$ 's and $\begin{array}{r}y\end{array}$  's.
  ::首先简化括号中的表达方式。 首先, 使用代表属性的权力的商数将 x 和 y 合并 。

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  $$\begin{array}{r}{\left(\frac{x^{\text{-}2}y}{x^4{y}^3}\right)}^{\text{-}2}={(x^{-6}{y}^{-2})}^{-2}\end{array}$$

  
  :x-2yx4y3)-2=(x-6y-2)-2-2

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  From here, focus on making the exponents positive. One option if to move any factors with negative exponents into the denominator. However,  using the power of products property to distribute the -2 as an exponent will cancel the negative exponents.
  ::从这里开始, 重点是让指数成为正数。 如果将负数指数的任何因素移入分母, 一个选项。 但是, 使用产品产权的力量来分配 - 2 指数将会取消负数指数 。

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  $$\begin{array}{r}{(x^{-6}{y}^{-2})}^{-2}=x^{(-6\times -2)}{y}^{(-2\times -2)}\end{array}$$

  
  :x-6y-2)-2=x(-6 2)y(-2 2)

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  Answer:  $\begin{array}{r}{x}^{12}{y}^4\end{array}$  
  ::答复:x12y4

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  Discussion Questions : 
  ::讨论问题:

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  1. What are some other ways in which you could have simplified the expression?
     ::还有什么其他方法可以简化表达方式?
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  2. Assign different values other than 0, 1, or 2 to  $\begin{array}{r}x\end{array}$  and  $\begin{array}{r}y\end{array}$  in the solution and the original expression and evaluate both expressions. What do you notice?
     ::指定除 0、 1、 2 或 2 到 x 和 y 以外的不同值,在解决方案中,在原始表达式和初始表达式中,并评价两个表达式。您注意到什么?

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  E xponent rules help describe the characteristics of extremely small or extremely large objects. Scientific  notation is used to write these small or large numbers. Exponent rules are used to manipulate these values since numbers written in scientific notation contain exponents.   
  ::指数规则有助于描述极小或极大物体的特性。 科学标记用于写这些小或大数字。 指数规则用于操纵这些数值, 因为科学标记中写的数字含有指数。

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  Example
  ::示例示例示例示例

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  A rectangular nano-chip has the dimensions of  $\begin{array}{r}a\times {10}^{-9}\end{array}$  by $\begin{array}{r}b\times {10}^{-9}\end{array}$   meters. Write an expression to represent the area of the chip.
  ::矩形纳米芯片的尺寸为ax10-9乘以 bx10-9米。写一个表达式以表示芯片的面积。

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  To find the area , multiply the dimensions of the chip.
  ::为了找到这个区域, 乘以芯片的维度。

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  $$\begin{array}{r}a\times {10}^{-9}\times b\times {10}^{-9}\end{array}$$

  
  ::ax10-9xxbx10-9

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  U se the commutative property to rearrange the expression.
  ::使用通量属性重新排列表达式。

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  $$\begin{array}{r}a\times {10}^{-9}\times b\times {10}^{-9}=a\times b\times {10}^{-9}\times {10}^{-9}\end{array}$$

  
  ::a_BAR__BAR___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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  Combining the powers of 10, will  result in the following:
  ::合并10国权力将产生以下结果:

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  $$\begin{array}{rl}a\times b\times {10}^{-9}\times {10}^{-9}& =a\times b\times {10}^{(-9+-9)}\\ & =a\times b\times {10}^{-18}\end{array}$$

  
  ::axbx10-9x10-9=axbxx10(-99)=axbx10-18

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  The factor  $\begin{array}{r}a\times b\end{array}$  represents the area of the nano-chip in nanometers. The value of  $\begin{array}{r}a\times b\end{array}$  will be in a unit called attometers  $\begin{array}{r}({10}^{-18}\text{ meters}).\end{array}$   This value  represents the unit of measure for the area in terms of square meters.  Similarly,  $\begin{array}{r}a\times {10}^{-9}\end{array}$  represents  $\begin{array}{r}a\end{array}$   nanometers with the  $\begin{array}{r}{10}^{-9}\end{array}$  converting those nanometers to meters.
  ::axb 表示纳米芯片在纳米计中的面积。 axb 的值将用一个单位单位(10-18米)表示,该数值以平方米表示面积的度量单位。同样, ax10-9 表示纳米计,以10- 9 将这些纳米计转换成米。

   

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  Discussion Question : How can you check to confirm that your simplified expression will be equivalent to the original expression?
  ::讨论问题:您如何检查确认您的简化表达式将等同于原表达式?

   

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  Activity 3: Rational Exponents
  ::活动3:理性指数

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  can be used to represent radicals. The numerator in a fractional exponent represents the power, and the denominator represents the root .
  ::可用于代表基体。 分数前列中的分子代表能量, 分母代表根。

  

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  Fractional exponents obey the same rules as the integer exponents explored thus far.
  ::分数前列者遵守与迄今所探索的整数前列者相同的规则。

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  Example
  ::示例示例示例示例

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  Simplify  $\begin{array}{r}{\left(81\right)}^{\text{-}\frac{1}{4}}\end{array}$
  ::简化(81-14)

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  For this example, the following exponent rules will be used:
  ::例如,将使用下列引言规则:

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    •   $\begin{array}{r}{a}^{\frac{n}{m}}=\sqrt[m]{a^n}\end{array}$
  ::• 上午=上午

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    •   $\begin{array}{rl}{a}^{-m}& =\frac{1}{a^m}\\ \end{array}$
  ::• 上午1时

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    •   $\begin{array}{r}{\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}\end{array}$
  ::• (ab)n=anbn

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  F irst, make the fractional exponent positive.
  ::首先,使分数前列正数为正数。

  $$\begin{array}{rl}{81}^{\text{-}\frac{1}{4}}& =\frac{1}{{81}^{\frac{1}{4}}}\end{array}$$

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  Next, apply the power of quotients exponent rule. Since 1 can be written as the base 1 to any power, use this to rewrite 1 as  $\begin{array}{r}{1}^{\frac{1}{4}}.\end{array}$
  ::下一步, 应用 商数 引言 规则 。 由于 1 可以 写为 1 基数 1 对任何权力 , 请使用它重写 1 114 。

  $$\begin{array}{rl}\frac{1}{{81}^{\frac{1}{4}}}=\frac{1^{\frac{1}{4}}}{{81}^{\frac{1}{4}}}& ={\left(\frac{1}{81}\right)}^{\frac{1}{4}}\end{array}$$

   

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  Finally ,  evaluate  the   fractional exponent using a radical . 
  ::最后,用激进来评估分数指数。

  $$\begin{array}{rl}{\left(\frac{1}{81}\right)}^{\frac{1}{4}}& =\sqrt[4]{\frac{1}{81}}\end{array}$$

   

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  Taking the fourth root of  $\begin{array}{r}\frac{1}{81}\end{array}$  will  produce   the answer.
  ::181的第四根将产生答案。

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  Answer:   $\begin{array}{r}\frac{1}{3}\end{array}$
  ::答复:13

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  If you stopped at  $\begin{array}{r}\frac{1}{{81}^{\frac{1}{4}}}=\frac{1}{\sqrt[4]{81}}\end{array}$   and rationalized it, would you have gotten the same answer? Use the questions  below to practice simplifying expressions with rational exponents.
  ::如果您在 18114 = 1814 停止并将其合理化, 您是否得到相同的答案 ? 使用下面的问题来用理性的推手来简化表达方式 。

   

   

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  Wrap-Up: Review Questions
  ::总结:审查问题

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     Summary
  ::摘要

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  When simplifying an exponential expression, the rules below must be followed. However, the order in which the rules are followed may vary based on the structure of the expression.
  ::在简化指数表达式时,必须遵守以下规则,但是,遵守规则的顺序可能因表达式的结构而不同。

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  • $\begin{array}{r}\frac{a^m}{a^n}=a^{m-n}\end{array}$  
    ::aman=am-n (aman=am-n)
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  • $\begin{array}{r}{a^m}^n=a^{mn}\end{array}$  
    ::AMN= AMN 显示器
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  • $\begin{array}{r}{a}^0=1\end{array}$  
    ::0.0=1
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  • $\begin{array}{r}{\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}\end{array}$  
    :ab)n=anbn
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  • $\begin{array}{r}{a}^{-m}=\frac{1}{a^m}\end{array}$  
    ::a-m=1毫米
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  • $\begin{array}{r}{a}^{\frac{m}{n}}=\sqrt[n]{a^m}\end{array}$  
    ::AMN= AMN 显示器