Section outline

  • Exponent of a Quotient 
    ::引数指数

    When we raise a whole quotient to a power, another special rule applies. Here is an example:
    ::当我们向权力提出一个整体商数时,适用另一个特殊规则。例如:

    ( x 3 y 2 ) 4 = ( x 3 y 2 ) ( x 3 y 2 ) ( x 3 y 2 ) ( x 3 y 2 ) = ( x x x ) ( x x x ) ( x x x ) ( x x x ) ( y y ) ( y y ) ( y y ) ( y y ) = x 12 y 8

    :sadx3xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx}}(y})

    Notice that the exponent outside the " data-term="Parentheses" role="term" tabindex="0"> parentheses is multiplied by the exponent in the numerator and the exponent in the denominator, separately. This is called the power of a quotient rule :
    ::注意括号外的引言方乘以分子中的引言方和分母中的引言方。这称为商数规则的力量:

    Power Rule for Quotients: ( x n y m ) p = x n p y m p
    ::引号规则 : (xnym) p=xnpymp

    Let’s apply these new rules to a few examples.
    ::让我们把这些新规则运用于几个例子。

    Simplifying Expressions 
    ::简化表达式

    1. Simplify the following expressions.
    ::1. 简化以下表达式。

    a)  4 5 4 2
    :sada) 4542

    We can use the quotient rule first and then evaluate the result: 4 5 4 2 = 4 5 2 = 4 3 = 64
    ::我们可以首先使用商数规则, 然后对结果进行评估: 4542=45-2=43=64

    OR we can evaluate each part separately and then divide: 4 5 4 2 = 1024 16 = 64
    ::我们可以分别评估每个部分,然后划分:4542=102416=64

    b)  5 3 5 7
    ::b) 5357

    Use the quotient rule first and then evaluate the result: 5 3 5 7 = 1 5 4 = 1 625
    ::首先使用商数规则, 然后评价结果: 5357=154=1625

    OR evaluate each part separately and then reduce: 5 3 5 7 = 125 78125 = 1 625
    ::或分别评价每一部分,然后减少:5357=12578125=1625

    Notice that it makes more sense to apply the quotient rule first for examples (a) and (b). Applying the exponent rules to simplify the expression before plugging in actual numbers means that we end up with smaller, easier numbers to work with.
    ::请注意,首先对例子(a)和(b)适用商数规则更有意义。在插入实际数字之前,应用引言规则简化表达式,意味着我们最终会遇到较小、更容易操作的数字。

    c)  ( 3 4 5 2 ) 2
    :sadc) (3452)2

    Use the power rule for quotients first and then evaluate the result: ( 3 4 5 2 ) 2 = 3 8 5 4 = 6561 625
    ::先对商数使用功率规则,然后再对结果进行评估sad3452)2=3854=6561625

    OR evaluate inside the parentheses first and then apply the exponent: ( 3 4 5 2 ) 2 = ( 81 25 ) 2 = 6561 625
    ::OR 首先在括号内评价, 然后应用表号sad3452)2=(8125)2=6561625

    2. Simplify the following expressions:
    ::2. 简化以下表述:

    a)  x 12 x 5
    ::a) x12x5

    b)  ( x 4 x ) 5
    :sadb) (x4x)5

    Use the power rule for quotients and then the quotient rule: ( x 4 x ) 5 = x 20 x 5 = x 15
    ::对商数使用功率规则, 然后对商数规则使用功率规则 sad x4x) 5=x20x5=x15

    OR use the quotient rule inside the parentheses first, then apply the power rule: ( x 4 x ) 5 = ( x 3 ) 5 = x 15
    ::OR 先在括号内使用商数规则, 然后应用权力规则 (x4x) 5= (x3)5=x15

    3. Simplify the following expressions.
    ::3. 简化以下表达式。

    When we have a mix of numbers and variables, we apply the rules to each number or each variable separately.
    ::当我们混合了数字和变量时,我们分别对每个数字或每个变量分别适用规则。

    a)  6 x 2 y 3 2 x y 2
    ::a) 6x2y32xy2

    Group like terms together: 6 x 2 y 3 2 x y 2 = 6 2 x 2 x y 3 y 2
    ::组合类似条件组 : 6x2y32xy2= 62xx2xy3y2

    Then reduce the numbers and apply the quotient rule on each fraction to get 3 x y .
    ::然后减少数字,对每个分数应用商数规则,以获得三轴。

    b)  ( 2 a 3 b 3 8 a 7 b ) 2
    :sadb) (2a3b38a7b)2

    Apply the quotient rule inside the parentheses first: ( 2 a 3 b 3 8 a 7 b ) 2 = ( b 2 4 a 4 ) 2
    ::首先在括号内应用商数规则sad2a3b38a7b)2=(b24a4)2

    Then apply the power rule for quotients: ( b 2 4 a 4 ) 2 = b 4 16 a 8
    ::然后对商数应用功率规则sadb24a4)2=b416a8

    Examples 
    ::实例

    Simplify the following expressions.
    ::简化下列表达式。

    In problems where we need to apply several rules together, we must keep the in mind.
    ::在我们需要共同适用若干规则的问题中,我们必须牢记这一点。

    Example 1
    ::例1

    ( x 2 ) 2 x 6 x 4
    :sadx2)2x6x4

    We apply the power rule first on the first term :
    ::我们首先运用第一任期的权力规则:

    ( x 2 ) 2 x 6 x 4 = x 4 x 6 x 4

    :sadx2) 2x6x4=x4x6x4

    Then apply the quotient rule to simplify the fraction:
    ::然后运用商数规则简化分数:

    x 4 x 6 x 4 = x 4 x 2

    ::x4x6x4=x4x2

    And finally simplify with the product rule:
    ::最后,用产品规则来简化:

    x 4 x 2 = x 6

    ::x4x2=x6

    Example 2
    ::例2

      ( 16 a 2 4 b 5 ) 3 b 2 a 16
    :sad16a24b5)3b2a16

    Simplify inside the parentheses by reducing the numbers:
    ::通过减少数字简化括号内的内容:

    ( 4 a 2 b 5 ) 3 b 2 a 16

    :sad4a2b5)、3b2a16

    Then apply the power rule to the first fraction:
    ::然后对第一个分数应用权力规则:

    ( 4 a 2 b 5 ) 3 b 2 a 16 = 64 a 6 b 15 b 2 a 16

    :sad4a2b5)3b2a16=64a6b15b2a16

    Group like terms together:
    ::组式相似的术语组合在一起 :

    64 a 6 b 15 b 2 a 16 = 64 a 6 a 16 b 2 b 15

    ::64a6b15b2a16=64a6a16b15

    And apply the quotient rule to each fraction:
    ::对每一分数应用商数规则:

    64 a 6 a 16 b 2 b 15 = 64 a 10 b 13

    ::=64a6a16b2b15=64a10b13

    Review 
    ::回顾

    Evaluate the following expressions.
    ::评估以下表达式。

    1. ( 3 8 ) 2
    2. ( 2 2 3 3 ) 3
    3. ( 2 3 4 2 2 4 ) 2

    Simplify the following expressions.
    ::简化下列表达式。

    1. ( a 3 b 4 a 2 b ) 3
      :sada3b4a2b)3
    2. ( 18 a 4 15 a 10 ) 4
      :sad18a415a10)4
    3. ( x 6 y 2 x 4 y 4 ) 3
      :sadx6Y2x4y4)3
    4. ( 6 a 2 4 b 4 ) 2 5 b 3 a
      :sad6a24b4)2
    5. ( 2 a 2 b c 2 ) ( 6 a b c 3 ) 4 a b 2 c
      :sad2a2bc2(6abc3)4ab2c)
    6. ( 2 a 2 b c 2 ) ( 6 a b c 3 ) 4 a b 2 c for a = 2 , b = 1 , and c = 3
      :sada=2,b=1,和c=3的2a2bc2(6abc3)4ab2c
    7. ( 3 x 2 y 2 z ) 3 z 2 x for x = 1 , y = 2 , and z = 1
      :sad3x2y2z) 3z2x x=1,y=2, z1
    8. 2 x 3 x y 2 ( x 2 y ) 2 for x = 2 , y = 3
      ::2x3xy2}(x2y) 2 x=2, y @% 3
    9. 2 x 3 x y 2 ( x 2 y ) 2 for x = 0 , y = 6
      ::x=0, y=6 2x3xy2}(x2y) 2 代表 x=0, y=6
    10. If a = 2 and b = 3 , simplify ( a 2 b ) ( b c ) 3 a 3 c 2 as much as possible.
      ::a=2和b=3,尽可能简化(a2(b)(b)(c)3a3c2)。

    Review (Answers)
    ::回顾(答复)

    Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
    ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。