Section outline

  • Gupta knows the area and width of a rectangle. He comes up with this equation for the length of the rectangle 2 x 2 1 2 x x + 1 . What is the length of the rectangle in simplified form?

    Complex Fractions
    ::复杂分数

    A complex fraction is a fraction that has fractions in the numerator and/or denominator. To simplify a complex fraction, you will need to combine all that you have learned about simplifying fractions in general.
    ::复数分数是指在分子和(或)分母中含有分数的分数。要简化一个复杂的分数,您需要将您所学到的关于一般简化分数的所有分数合并在一起。

    Let's simplify the following complex fractions.
    ::让我们简化以下复杂的部分。

    1. 9 x x + 2 3 x 2 4
      ::9xx+23x2-4

    Rewrite the complex fraction as a division problem.
    ::将复杂分数重写为分裂问题 。

    9 x x + 2 3 x 2 4 = 9 x x + 2 ÷ 3 x 2 4

    ::9xx+23x2 - 4=9xx+2}3x2 - 4

    Flip the second fraction, change the problem to multiplication and simplify.
    ::翻转第二段,将问题改为乘法和简化。

    9 x x + 2 ÷ 3 x 2 4 = 9 x x + 2 x 2 4 3 = 9 x 3 x + 2 ( x + 2 ) ( x 2 ) 3 = 3 x ( x 2 )

    ::9xx+23x2-4=9xx+2x2x2-43=9x3xxx2}(x+2)(x-2)(x-2)3=3x(x-2)

    1. 1 x + 1 x + 1 4 1 x
      ::1x1+1x+14-1x

    To simplify this complex fraction, we first need to add the fractions in the numerator and subtract the two in the denominator. The LCD of the numerator is x ( x + 1 ) and the denominator is just x .
    ::为了简化这一复杂分数, 我们首先需要在分子中添加分数, 并在分母中减去这两个分数。 分子的LCD是 x( x+1) , 分数只是 x 。

    1 x + 1 x + 1 4 1 x = x + 1 x + 1 1 x + 1 x + 1 x x x x 4 1 x = x + 1 x ( x + 1 ) + x x ( x + 1 ) 4 x x 1 x = 2 x + 1 x ( x + 1 ) 4 x 1 x

    ::1x+1x1x+14-1x=1x1x1x1x1x1x1x1x1x1x1xxxxxxxx1xxxx1xxx1x1x1x1x1xxxxxxxx+1x1x1x1x1x1xxx1x1x1x1x1xx1x1x1xx1xx1x1x1xx1xxx1xxx1xxx1xx1xxx1xxxxxx1xxx1xxx1x1x1x1x1x1xxxx1xxx1x1x1x1x1x1x1x1x1x1xxx1xx1x1xxxx1x1x1x1xx1x1x1x1x1xx1x1x1x1x1x1x1xx1x1x1xx1x1x1x1xxx1x1xx1x1xxx1x1x1x1xx1xx1xx1xxxx1x1x1xx1xxx1xxxx1xxxxx1xxxx1xxxxxxxxxxx1xxxxxxxxxxx1xxxxxxxxxxxxx1xxxxxxx1xxxxxxxx1xxxxxxxxxxxxxxxxxxxxxxx

    Divide and simplify if possible.
    ::在可能的情况下,分开和简化。

    2 x + 1 x ( x + 1 ) 4 x 1 x = 2 x + 1 x ( x + 1 ) ÷ 4 x 1 x = 2 x + 1 x ( x + 1 ) x 4 x 1 = 2 x + 1 ( x + 1 ) ( 4 x 1 )

    ::2x+1x( x+1) 4x -1x=2x1x1x( x+1) 4x -1x=2x1x( x+1) x4x -1x=2x1x( x+1) x4x -1x=2x1x1( x+1) (4x-1)

    1.   5 x x 2 + 6 x + 8 + x x + 4 6 x + 2 2 x + 3 x 2 3 x 10
      ::5-2x2+6x+8+6x8+xxxxx+46xx+2-2-2x+3x2-3x2-3xx-10

    First, add the fractions in the numerator and subtract the ones in the denominator.
    ::首先,在分子中添加分数,并在分母中减去分数。

    5 x x 2 + 6 x + 8 + x x + 4 6 x + 2 2 x + 3 x 2 3 x 10 = 5 x ( x + 4 ) ( x + 2 ) + x x + 4 x + 2 x + 2 x 5 x 5 6 x + 2 2 x + 3 ( x + 2 ) ( x 5 ) = 5 x + x ( x + 2 ) ( x + 4 ) ( x + 2 ) 6 ( x 5 ) ( 2 x + 3 ) ( x + 2 ) ( x 5 ) = x 2 + x + 5 ( x + 4 ) ( x + 2 ) 4 x 36 ( x + 2 ) ( x 5 )

    ::5 -xx2+6x8+6x8+6x8+xx2+46x+2x2+2x2+2x2x2+2x2x2x3x3x3x3xxx10=5x(x+4)(x+2)+x2x+2xx-2x5x5x5x2x5x2x5x6x2+2x6x+6x6x+6x6x+6x6x+2x2x2x2x2xx2xx2xx2x2xx2x6x+5x2xx5xx2x4x+4x2x5x5x2x5xxxx2x2x6xxx5x2x5x+2x2x(x+2x2x)(x+2x)(x-5)6x5xxxxxx3x3x3x3x(x2+2)(x5)

    Now, rewrite as a division problem, flip, multiply, and simplify.
    ::现在,重写为分裂问题,翻转,乘法和简化。

    x 2 + x + 5 ( x + 4 ) ( x + 2 ) 4 x 36 ( x + 2 ) ( x 5 ) = x 2 + x + 5 ( x + 4 ) ( x + 2 ) ÷ 4 x 36 ( x + 2 ) ( x 5 ) = x 2 + x + 5 ( x + 4 ) ( x + 2 ) ( x + 2 ) ( x 5 ) 4 ( x 9 )

    ::x2+x+2(x+4)(x+2)(x+2)(x-5)=x2+5(x+4)(x+2)(x+2)**4x-36(x+2)(x-5)=x2+5(x+4)(x+4)(x+2)(x+2)(x+2)__(x+2)(x+2)(x+2)(x-2)(x-5)4(x-9)

    = ( x 2 + x + 5 ) ( x 5 ) 4 ( x + 4 ) ( x 9 )


    ::=(x2+x+5)(x-5)(x-5)4(x+4)(x-9)

    Examples
    ::实例

    Example 1
    ::例1

    Earlier, you were asked to determine  the length of a given  rectangle in simplified form. 
    ::早些时候,有人要求你以简化的形式确定给定矩形的长度。

    Rewrite the complex fraction as a division problem.
    ::将复杂分数重写为分裂问题 。

    2 x 2 1 2 x x + 1 = 2 x 2 1 ÷ 2 x x + 1

    ::2x2 - 12xxx+1=2x2 - 1=2x2 - *2xxx+1

    Flip the second fraction, change the problem to multiplication and simplify.
    ::翻转第二段,将问题改为乘法和简化。

    2 x 2 1 ÷ 2 x x + 1 = 2 x 2 1 x + 1 2 x = 2 ( x + 1 ) ( x 1 ) ( x + 1 ) 2 x = 1 x 2 x

    ::2x2 - 1 - 1x2 - 2xx2+1=2x2 - 1xxxx+12x=2(x+1)(x-1)(x-1)(x+1)(x+1)__(x+1)_(x+1)2x=1x2-x

    Therefore , the length of the rectangle in simplified form is 1 x 2 x .
    ::因此,简化形式的矩形长度为1x2-x。

    S implify the complex fractions.
    ::简化复杂的分数 。

    Example 2
    ::例2

    5 x 20 x 2 x 4 x
    ::5x-20x2x-4x

    Rewrite the fraction as a division problem and simplify.
    ::将分数重写为分裂问题并简化 。

    5 x 20 x 2 x 4 x = 5 x 20 x 2 ÷ x 4 x = 5 ( x 4 ) x 2 x x 4 = 5 x
    ::5x - 20x2x - 4x=5x - 20x2x - 4x=5(x-4)x2x - 4x=5x

    Example 3
    ::例3

    1 x x 2 x 1 1 + 1 x
    ::1-xx-2-2x-11+1x

     Add the fractions in the numerator and denominator together.
    ::在分子和分母中同时添加分数。

    1 x x 2 x 1 1 + 1 x = x 1 x 1 1 x x 2 x 1 x x x x 1 + 1 x = ( x 1 ) ( 1 x ) 2 x x ( x 1 ) x + 1 x = x 2 + 1 x ( x 1 ) x + 1 x
    ::1-xxxxxx-1-1xx+1x=x-1x-1x-1x-11x-1=x-2x-1x-1xxxxxxxxxx-1xxx-1(1-xx)-2xxx(x-1-1)(1-x)-2xx(x-1)xxxx+1xx(x-1)xxxxxxxxxxxxxxxx2+1xx(x-1)xxx+1xxxxxxx

    Now, rewrite the fraction as a division problem and simplify.
    ::现在,将分数重写为分数问题并简化 。

    x 2 + 1 x ( x 1 ) ÷ x + 1 x = ( x 2 1 ) x ( x 1 ) x x + 1 = ( x 1 ) ( x + 1 ) x ( x 1 ) x x + 1 = 1

    ::-x2+1x(x- 1) x+1x(x2- 1)xxxxxxxx(x1)xxxxxx+1}(x-1)xxx(x-1)xx(x-1)xxxxxxxxxx+1*1

    Example 4
    ::例4

      - 3 - 4 x 2 5 x + 6 + - 4 - 4 x + 3 - 1 - 4 x 2 5 x + 6 + 2 - 4 x + 3
    ::-3-4x2-5x+6+4-4x+3-1-4x2-5x+6+2-4x+3

    Add the numerator and the denominator of this complex fraction.
    ::添加此复杂分数的分子和分母。

    - 3 - 4 x 2 5 x + 6 + - 4 - 4 x + 3 - 1 - 4 x 2 5 x + 6 + 2 - 4 x + 3 = - 3 ( - 4 x + 3 ) ( x + 2 ) + ( - 4 ) ( - 4 x + 3 ) ( x + 2 ) ( x + 2 ) - 1 ( - 4 x + 3 ) ( x + 2 ) + 2 - 4 x + 3 ( x + 2 ) ( x + 2 ) = - 4 x 11 - 4 x 2 5 x + 6 2 x + 3 - 4 x 2 5 x + 6 = - 4 x 11 - 4 x 2 5 x + 6 - 4 x 2 5 x + 6 2 x + 3 = - 4 x 11 2 x + 3

    ::- 3-4x2-5x6+6x6(x+2)-5x+4+4+3-4-4x2-5x6+6+2-4-4x3+3=-3(-4x+3)(x+2+3)+2(4x+3)+(4x+3)+(x+2)×-1(x+2)(x+2)(x+2)+2)+2-4x3x3+3(x2)(x+2)(x+2)(4x+3)(x+2)+2)+2+2+2+2-44x+3(x+2)(x+2)(x+2)(x+3)(x+3)(x+2)+2+2+2+2+2+2+2-44x+3(x+2)(x+2)(x+2)(x+2)(x+2)(x)(x+2)=-4-4x+3x+3x+3(x)+(x+2)=4x-11-44x2-4x2-4x2x2x2x2x-2-5x2=5x2=+5x=+5x+5x+5x+6x+6x+6x+6x+6x+6x+6x+6x6x6x6x+6x6x6x6x6x6x+6x6x6x6x6x6x6x+4x6x6x=4x5x=4x=4x5x5x5x5x5x=4x=4x=4x=4x=4x=4x=4x=4x=4x=4x=4x=4xxxxx4xxxxxxxxxxxxxxx4x4x4x4x4x4x4x4x4x4x4x4x4x4x4x4x4x=4x=4x=4x=4xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

    Review
    ::回顾

    Simplify the complex fractions.
    ::简化复杂的分数 。

    1. 2 x 5 8 7
      ::2x587
    2. 4 x 2 9 6 x x + 3
      ::4x2-96xxx+3
    3. 7 x 3 x 2 + 5 x + 6 35 x 2 x + 2
      ::7x3x2+5x+635x2x+2
    4. 24 x + 3 3 x + 1 16 x + 2 6 x 2 13 x 5
      ::24x+33x+116x+26x2-13x-5
    5. 4 x 1 + 1 x 1 x 5
      ::4-1+1x1x-5
    6. 3 x x + 4 1 x 3 x 4 x 2 + 6 x + 8
      ::3x+4-1x3x-4x2+6x+8
    7. 8 3 x x + 5 10 x + 5 + 5 x + 1
      ::8-3xx+510x+5+5x+5x+1
    8. x x + 3 4 2 x + 1 3 2 x + 1 + 6 x 2 9
      ::xx+3-42x+132x+1+6x2-9
    9. x + 3 x + 2 x 5 x 3 2 x 4 x x 5
      ::x+3x+2x5-x32x-4xxx-5
    10. 2 x 5 x 2 13 x 6 + 1 x 3 4 5 x + 2 5 x 5 x 2 3 x 2
      ::2x5x2-13x-6+1x-345x+2-5x5x2-3x-2
    11. 3 x x 2 4 + x + 4 x 2 + 3 x + 2 x + 1 x 2 x 2 2 x x 2 + 2 x + 1
      ::3xx2-4+x4x2+3x2x2+2x1x2-x-2-2x2+2x2+2x1+1

    Use the following pattern to answer the next four questions.
    ::使用以下模式回答下四个问题。

    2 + 1 1 + 1 2 ,   2 + 1 1 + 1 2 + 2 3 ,   2 + 1 1 + 1 2 + 2 3 + 3 4

    1. Find the next two terms in the pattern.
      ::查找图案中的下两个术语。
    2. Using your graphing calculator, simplify each term in the pattern to a decimal.
      ::使用您的图形计算计算器,将模式中的每个词简化为小数点后的一个词。
    3. Make a conjecture about this pattern and the number the terms appear to be approaching.
      ::对这种模式和术语似乎即将到来的数量进行猜测。
    4. Find the sixth term in the pattern. Does it support your conjecture?
      ::查找模式中的第六学期。 它是否支持您的猜测 ?

    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 。