课程： 12.7 逻辑表达式的排除值 | The Way To Learn

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逻辑表达式的排除值
Excluded Values for Rational Expressions
::逻辑表达式的排除值

A simplified rational expression is one where the numerator and denominator have no common factors. In order to simplify an expression to lowest terms , we factor the numerator and denominator as much as we can and cancel common factors from the numerator and the denominator.
::简化的理性表达法是分子和分母没有共同因素的表达法。为了将表达法简化为最低条件, 我们尽可能对分子和分母进行分解, 从分子和分母中取消共同因素。

Simplifying Rational Expressions
::简化逻辑表达式

Reduce each rational expression to simplest terms.
::将每种合理表达方式减为最简单的表达方式。

a) $\begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*}$
:a) 4x-22x2+x-1

$\begin{aligned} Factor the numerator and denominator completely: \frac{2 (2 x - 1)}{(2 x - 1) (x + 1)} \\ Cancel the common factor (2 x - 1) : \frac{2}{x + 1} \end{aligned}$
$\begin{align*}\text{Factor the numerator and denominator completely:} \qquad \frac{2(2x-1)}{(2x-1)(x+1)}\!\\ \\ \text{Cancel the common factor} \ (2x - 1): \qquad \qquad \qquad \qquad \qquad \frac{2}{x+1}\end{align*}$
::完全乘以分子和分母 : 2 2x- 1 (2x- 1) - 1 (x+1) - 1 (x+1) 。

b) $\begin{align*}\frac{x^2-2x+1}{8x-8}\end{align*}$
:b) x2-2x+18x-8

$\begin{aligned} Factor the numerator and denominator completely: \frac{(x - 1) (x - 1)}{8 (x - 1)} \\ Cancel the common factor (x - 1) : \frac{x - 1}{8} \end{aligned}$
$\begin{align*}\text{Factor the numerator and denominator completely:} \qquad \frac{(x-1)(x-1)}{8(x-1)}\!\\ \\ \text{Cancel the common factor}\ (x - 1): \qquad \qquad \qquad \qquad \qquad \ \ \frac{x-1}{8}\end{align*}$
::完全乘以分子和分分母: (x- 1)(x- 1)(x- 1)8(x-1)8(x- 1)C

c) $\begin{align*}\frac{x^2-4}{x^2-5x+6}\end{align*}$
:c) x2 - 4x2 - 5x+6

$\begin{aligned} Factor the numerator and denominator completely: \frac{(x - 2) (x + 2)}{(x - 2) (x - 3)} \\ Cancel the common factor (x - 2) : \frac{x + 2}{x - 3} \end{aligned}$
$\begin{align*}\text{Factor the numerator and denominator completely:} \qquad \frac{(x-2)(x+2)}{(x-2)(x-3)}\!\\ \\ \text{Cancel the common factor} (x - 2): \qquad \qquad \qquad \qquad \qquad \quad \frac{x+2}{x-3}\end{align*}$
::完全乘以分子和分母(x- 2)(x+2)(x-2)(x-2)(x-3)(x-3)(x-3)(x-3)(x-2):x+2x-3)

When reducing fractions, you are only allowed to cancel common factors from the denominator but NOT common terms. For example, in the expression $\begin{align*}\frac{(x+1) \cdot (x-3)}{(x+2) \cdot (x-3)}\end{align*}$ , we can cross out the $\begin{align*}(x - 3)\end{align*}$ factor because $\begin{align*}\frac{(x-3)}{(x-3)}=1\end{align*}$ . But in the expression $\begin{align*}\frac{x^2+1}{x^2-5}\end{align*}$ we can’t just cross out the $\begin{align*}x^2\end{align*}$ terms.
::当减小分数时, 您只能从分母中取消共同因数, 但非常见因数。例如, 在表达式 (x+1) {(x- 3)(x+2) {(x- 3) }}} (x- 3) } (x- 3) } (x-3) =1) 。但在表达式 x2+1x2 - 5 中, 我们无法直接跳出 x2 术语。

Why can’t we do that? When we cross out terms that are part of a sum or a difference , we’re violating the (PEMDAS). Remember, the fraction bar means division . When we perform the operation $\begin{align*}\frac{x^2+1}{x^2-5}\end{align*}$ , we’re really performing the division $\begin{align*}(x^2+1) \div (x^2-5)\end{align*}$ — and the order of operations says that we must perform the operations inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses before we can perform the division.
::为什么我们不能这样做呢?当我们跨出属于一个总和或一个差异的部分条件时,我们违反了(PEMDAS ) 。记住, 分数条意味着分割。当我们执行x2+1x2-5, 我们真的执行分割(x2+1)\\\(x2-5) —— 操作顺序表明我们必须在括号内执行操作,然后才能执行分割。

Using numbers instead of variables makes it more obvious that canceling individual terms doesn’t work. You can see that $\begin{align*}\frac{9+1}{9-5}=\frac{10}{4}=2.5\end{align*}$ — but if we canceled out the 9’s first, we’d get $\begin{align*}\frac{1}{-5}\end{align*}$ , or -0.2, instead.
::使用数字而不是变量更明显地说明取消个别条件行不通。你可以看到9+19—5=104=2.5 — — 但如果我们先取消9个数字,我们就会得到1—5或-0.2。

Finding Excluded Values of Rational Expressions
::寻找排除的逻辑表达式值

Whenever there’s a variable expression in the denominator of a fraction , we must remember that the denominator could be zero when the independent variable takes on certain values. Those values, corresponding to the vertical asymptotes of the function , are called excluded values. To find the excluded values, we simply set the denominator equal to zero and solve the resulting equation .
::当分数分母存在变量表达式时,我们必须记住当独立变量吸收某些值时分母可能是零。这些值与函数的垂直空位相对应,被称作排除值。要找到被排除的值,我们只需将分母设为零,并解决由此产生的方程式。

Find the excluded values of the following expressions.
::查找下列表达式的排除值。

a) $\begin{align*}\frac{x}{x+4}\end{align*}$
::a)xx+4

$\begin{aligned} When we set the denominator equal to zero we obtain: x + 4 = 0 \Rightarrow x = - 4 \\ So - 4 is the excluded value. \end{aligned}$
$\begin{align*}\text{When we set the denominator equal to zero we obtain:} \quad \ \ x+4=0 \Rightarrow x=-4\!\\ \\ \text{So} \ \mathbf{-4} \ \text{is the excluded value.}\end{align*}$
::当我们设定等于零的分母时,我们获得的分母为: x+4=0x4So- 4是排除值。

b) $\begin{align*}\frac{2x+1}{x^2-x-6}\end{align*}$
::b) 2x+1x2-x-6

$\begin{aligned} When we set the denominator equal to zero we obtain: x^{2} - x - 6 = 0 \\ Solve by factoring: (x - 3) (x + 2) = 0 \\ \Rightarrow x = 3 and x = - 2 \\ So 3 a n d - 2 are the excluded values. \end{aligned}$
$\begin{align*}\text{When we set the denominator equal to zero we obtain:} \qquad x^2-x-6=0\!\\ \\ \text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \qquad (x-3)(x+2)=0\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \Rightarrow x=3 \ \text{and}\ x = -2\!\\ \\ \text{So}\ \mathbf{3}\ \mathbf{and}\ \mathbf{-2} \ \text{are the excluded values.}\end{align*}$
::当我们设定等于零的分母时,我们获得的分母为:x2-x-6=0,乘以乘数x-3)(x+2)=0 x=3 和 x2So 3 和-2)为排除值。

Removable Zeros
::可移动零

Removable zeros are those zeros from the original expression, but is not a zero for the simplified version of the expression. However, we have to keep track of them, because they were zeros in the original expression. This is illustrated in the following examples.
::可移除的零是原表达式的零,但该表达式的简化版本不是零。然而,我们必须跟踪这些零,因为它们在原表达式中是零。下面的例子说明了这一点。

Determining Removable Values
::确定可移动值

1. Determine the removable values of $\begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*}$ .
::1. 确定4x-22x2+x-1的可移动值。

Notice that in the expressions in Example A, we removed a division by zero when we simplified the problem. For instance, we rewrote $\begin{align*}\frac{4x-2}{2x^2+x-1}\end{align*}$ as $\begin{align*}\frac{2(2x-1)}{(2x-1)(x+1)}\end{align*}$ . The denominator of this expression is zero when $\begin{align*}x = \frac{1}{2}\end{align*}$ or when $\begin{align*}x = -1\end{align*}$ .
::请注意,在例A中的表达式中,当我们简化问题时,我们删除了零除法。例如,我们将4x-22x2+x-1重写为2(2x-1)(2x-1)(2x-1)(x-1)(x+1)。当 x=12 或 x=1时,该表达式的分母为零。

However, when we cancel common factors, we simplify the expression to $\begin{align*}\frac{2}{x+1}\end{align*}$ . This reduced form allows the value $\begin{align*}x = \frac{1}{2}\end{align*}$ , so $\begin{align*}x = -1\end{align*}$ is its only excluded value.
::然而,当我们取消共同因素时,我们将表达式简化为 2x+1 。这样减小的表达式允许值 x=12, 所以 x1 是它唯一的排除值。

Technically the original expression and the simplified expression are not the same. When we reduce a radical expression to its simplest form, we should specify the removed excluded value. In other words, we should write our final answer as $\begin{align*}\frac{4x-2}{2x^2+x-1}=\frac{2}{x+1}, x \neq \frac{1}{2}\end{align*}$ .
::从技术上讲,最初的表达式和简化的表达式并不相同。当我们将激进表达式减为最简单的表达式时,我们应该指定被删除的排除值。换句话说,我们应该以 4x- 22x2+x- 1=2x+1=1x1,x**12 写入我们的最后答案。

2. Determine the removable values of the expressions from parts b and c of the first example.
::2. 确定第一个示例b和c部分中的表达式的可移动值。

We should write the answer from the first example, part b as $\begin{align*}\frac{x^2-2x+1}{8x-8}=\frac{x-1}{8}, x \neq 1\end{align*}$ .
::我们应该写第一个例子的答案, b部分为 x2 - 2x+18x-8=x - 18,x1。

The answer from the first example, part c as $\begin{align*}\frac{x^2-4}{x^2-5x+6}=\frac{x+2}{x-3}, x \neq 2\end{align*}$ .
::第一个例子的答案, c 部分为 x2 - 4x2 - 5x+6=x+2x-3,x @%2。

Example
::示例示例示例示例

Example 1
::例1

Find the excluded values of $\begin{align*}\frac{4}{x^2-5x}\end{align*}$ .
::查找4x2-5x的排除值。

$\begin{aligned} When we set the denominator equal to zero we obtain: x^{2} - 5 x = 0 \\ Solve by factoring: x (x - 5) = 0 \\ \Rightarrow x = 0 and x = 5 \\ So 0 a n d 5 are the excluded values. \end{aligned}$
$\begin{align*}\text{When we set the denominator equal to zero we obtain:} \quad \ \ x^2-5x=0\!\\ \\ \text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x(x-5)=0\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \Rightarrow x=0 \ \text{and} \ x = 5\!\\ \\ \text{So} \ \mathbf{0 \ and \ 5}\ \text{are the excluded values.}\end{align*}$
::当我们设定等于零的分母时,我们获得的分母为:x2-5x=0Solve,乘以乘数: x(x-5)=0 x=0和 x=5So 0和5是排除值。

Review
::回顾

Reduce each fraction to lowest terms.
::将每一分数减到最低值。

$\begin{align*}\frac{4}{2x-8}\end{align*}$
::42x-8

$\begin{align*}\frac{x^2+2x}{x}\end{align*}$
::x2+2xxx

$\begin{align*}\frac{9x+3}{12x+4}\end{align*}$
::9x+312x+4

$\begin{align*}\frac{6x^2+2x}{4x}\end{align*}$
::6x2+2x4x

$\begin{align*}\frac{x-2}{x^2-4x+4}\end{align*}$
::x-2x2 - 4x+4 x-2x2 - 4x+4

$\begin{align*}\frac{x^2-9}{5x+15}\end{align*}$
::x2-95x+15 x2-95x+15

$\begin{align*}\frac{x^2+6x+8}{x^2+4x}\end{align*}$
::x2+6x+8x2+4xx

$\begin{align*}\frac{2x^2+10x}{x^2+10x+25}\end{align*}$
::2x2+10xx2+10xx2+10x2+10xx+25

$\begin{align*}\frac{x^2+6x+5}{x^2-x-2}\end{align*}$
::x2+6x+5x2 -x-2

$\begin{align*}\frac{x^2-16}{x^2+2x-8}\end{align*}$
::x2 - 16x2+2x-8

$\begin{align*}\frac{3x^2+3x-18}{2x^2+5x-3}\end{align*}$
::3x2+3x-182x2+5x-3

$\begin{align*}\frac{x^3+x^2-20x}{6x^2+6x-120}\end{align*}$
::x3+x2 - 20x6x2+6x- 120

Find the excluded values for each rational expression.
::查找每个合理表达式的排除值。

$\begin{align*}\frac{2}{x}\end{align*}$
::2x 2x

$\begin{align*}\frac{4}{x+2}\end{align*}$
::4x+2 4x+2

$\begin{align*}\frac{2x-1}{(x-1)^2}\end{align*}$
::2x-1(x-1)2

$\begin{align*}\frac{3x+1}{x^2-4}\end{align*}$
::3x+1x2-4

$\begin{align*}\frac{x^2}{x^2+9}\end{align*}$
::x2x2+9

$\begin{align*}\frac{2x^2+3x-1}{x^2-3x-28}\end{align*}$
::2x2+3x-1x2-3x-28

$\begin{align*}\frac{5x^3-4}{x^2+3x}\end{align*}$
::5x3 - 4x2+3x

$\begin{align*}\frac{9}{x^3+11x^2+30x}\end{align*}$
::9x3+11x2+30x

$\begin{align*}\frac{4x-1}{x^2+3x-5}\end{align*}$
::4 - 1x2+3x-5

$\begin{align*}\frac{5x+11}{3x^2-2x-4}\end{align*}$
::5x+113x2-2-2x-4

$\begin{align*}\frac{x^2-1}{2x^2+x+3}\end{align*}$
::x2 - 12x2+x+3 x2+3

$\begin{align*}\frac{12}{x^2+6x+1}\end{align*}$
::12x2+6x+1

In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance. $\begin{align*}\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}\end{align*}$ . If $\begin{align*}R_1 = 25 \ \Omega\end{align*}$ and the total resistance is $\begin{align*}R_c = 10 \ \Omega\end{align*}$ , what is the resistance $\begin{align*}R_2\end{align*}$ ?
::在电路中,完全抗力的对等等于每次抗力的对等之和。 1Rc=1R1+1R2.如果R1=25 }和总抗力是Rc=10 ×,抵抗力是多少R2?

Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
::假设两个对象以20牛顿的引力相互吸引。如果两个对象之间的距离加倍,两个对象之间的新吸引力是什么?

Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
::假设两个对象以36个牛顿的引力相互吸引。如果两个对象的重量翻倍,如果两个对象之间的距离翻倍,那么两个对象之间的新的吸引力是什么?

A sphere with radius $\begin{align*}R\end{align*}$ has a volume of $\begin{align*}\frac{4}{3} \pi R^3\end{align*}$ and a surface area of $\begin{align*}4 \pi R^2\end{align*}$ . Find the ratio of the surface area to the volume of a sphere.
::半径为R的球体体体积为43R3,表面面积为4R2,找出表面积与球体体体积的比率。

The side of a cube is increased by a factor of 2. Find the ratio of the old volume to the new volume.
::立方体的侧面增加乘以 2 。查找旧音量与新音量的比例。

The radius of a sphere is decreased by 4 units. Find the ratio of the old volume to the new volume.
::球的半径减少 4 个单位。查找旧音量与新音量的比率。

Review (Answers)
::回顾(答复)

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::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。

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