课程： 12.12 使用比例比例的理性衡平 | The Way To Learn

The Way To Learn

首页日程管理

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节使用比例比例的衡平法

折叠展开
使用比例比例的衡平法
Rational Equations Using Proportions
::使用比例比例的衡平法

A rational equation is one that contains rational expressions. It can be an equation that contains rational coefficients or an equation that contains rational terms where the variable appears in the denominator.
::理性方程式包含合理的表达方式。它可以是包含合理系数的方程式,也可以是包含变量出现在分母中的合理条件的方程式。

An example of the first kind of equation is: $\begin{aligned} \frac{3}{5} x + \frac{1}{2} = 4 \end{aligned}$ .
::第一种等式的一个例子是:35x+12=4。

An example of the second kind of equation is: $\begin{aligned} \frac{x}{x - 1} + 1 = \frac{4}{2 x + 3} \end{aligned}$ .
::第二种等式的一个例子是:xx-1+1=42x+3。

The first aim in solving a rational equation is to eliminate all denominators. That way, we can change a rational equation to a polynomial equation which we can solve with the methods we have learned this far.
::解决理性等式的首要目标是消除所有分母。这样,我们可以将理性等式改变为多元等式,用我们迄今为止学到的方法解决。

Solve Rational Equations Using Cross Products
::使用交叉产品进行理理衡等同

A rational equation that contains just one term on each side is easy to solve by cross multiplication . Consider the following equation:
::每边只包含一个术语的合理方程式很容易通过乘法解决。考虑以下方程式 :

$\begin{aligned} \frac{x}{5} = \frac{x + 1}{2} \end{aligned}$

::x5=x+12 x5=x+12

Our first goal is to eliminate the denominators of both rational expressions. In order to remove the 5 from the denominator of the first fraction , we multiply both sides of the equation by 5:
::我们的第一个目标是消除两种理性表达的分母。为了从第一组分母的分母中去除5分母,我们要将等式的两边乘以5:

$\begin{aligned} 5 \cdot \frac{x}{5} & = 5 \cdot \frac{x + 1}{2} \\ x & = \frac{5 (x + 1)}{2} \end{aligned}$

::5x5=5xx+12x=5(x+1)2

Now, we remove the 2 from the denominator of the second fraction by multiplying both sides of the equation by 2:
::现在,我们从第二分数分数的分母中去掉2, 将方程的两边乘以2:

$\begin{aligned} 2 \cdot x & = 2 \cdot \frac{5 (x + 1)}{2} \\ 2 x & = 5 (x + 1) \end{aligned}$

::2x=25(x+1)22x=5(x+1)

Then we can solve this equation for $\begin{aligned} x \end{aligned}$ .
::然后我们就可以解开x的这个方程式

Notice that this equation is what we would get if we simply multiplied each numerator in the original equation by the denominator from the opposite side of the equation. It turns out that we can always simplify a rational equation with just two terms by multiplying each numerator by the opposite denominator; this is called cross multiplication.
::请注意,如果我们简单地将每个分子在原始方程中的分母乘以方程的对面的分母,这个方程就是我们能得到的。事实证明,我们总是可以通过将每个分子乘以对立分母,用两个条件来简化一个理性方程; 这叫做交叉乘法。

Solving for Unknown Values
::解决未知值

Solve the equation $\begin{aligned} \frac{2 x}{x + 4} = \frac{5}{x} \end{aligned}$ .
::解析方程 2xx+4=5x 。

$\begin{aligned} Cross-multiply. The equation simplifies to: & 2 x^{2} = 5 (x + 4) \\ Simplify: & 2 x^{2} = 5 x + 20 \\ Move all terms to one side of the equation: & 2 x^{2} - 5 x - 20 = 0 \\ Solve using the quadratic formula: & x = \frac{5 \pm \sqrt{185}}{4} \Rightarrow \underline{\underline{x = - 2.15}} or \underline{\underline{x = 4.65}} \end{aligned}$

::横乘。方程式简化为: 2x2=5( x+4) 简化为: 2x2=5xx+20 将所有条件都移动到方程式的一边: 2x2- 5x- 20=0Solve 使用四方形公式: x= 51854\\\\\\\\\\\\\\\\2.15_ 或 x=4. 65_ 或 x= 4. 65_ 。

It’s important to plug the answer back into the original equation when the variable appears in any denominator of the equation, because the answer might be an excluded value of one of the rational expressions. If the answer obtained makes any denominator equal to zero, that value is not really a solution to the equation.
::当变量出现在公式的任何分母中时,必须将答案塞回原来的方程,因为答案可能是理性表达的一种排除值。如果得到的答案使任何分母等于零,那么该数值实际上不是公式的解决方案。

Check: $\begin{aligned} \frac{2 x}{x + 4} = \frac{5}{x} \Rightarrow \frac{2 (- 2.15)}{- 2.15 + 4} \overset{?}{=} \frac{5}{- 2.15} \Rightarrow \frac{- 4.30}{1.85} \overset{?}{=} - 2.3 \Rightarrow - 2.3 = - 2.3. \end{aligned}$ The answer checks out.
::检查:2xx+4=5xx=5x=2(2-15)-2.15+4=?5-2.15=2.15=4.301.85=?-2.3=2.3=2.3}2.3=2.3。检查答案。

$\begin{aligned} \frac{2 x}{x + 4} = \frac{5}{x} \Rightarrow \frac{2 (4.65)}{4.65 + 4} \overset{?}{=} \frac{5}{4.65} \Rightarrow \frac{9.3}{8.65} \overset{?}{=} 1.08 \Rightarrow 1.08 = 1.08. \end{aligned}$ The answer checks out.
::2xx+4=5x2(4.654.65.65+4=?54.65=9.38.65=?1.08=1.08. 答案检查出来。

Solve Rational Equations Using Lowest Common Denominators
::使用最低的常见点

Another way of eliminating the denominators in a rational equation is to multiply all the terms in the equation by the lowest common denominator. You can use this method even when there are more than two terms in the equation.
::在合理等式中消除分母的另一种方法是将方程式中的所有条件乘以最低共同分母。即使方程式中有两个以上的条件,也可以使用这种方法。

1. Solve $\begin{aligned} \frac{3}{x + 2} - \frac{4}{x - 5} = \frac{2}{x^{2} - 3 x - 10} \end{aligned}$ .
::1. 解决 3x+2-4x-5=2x2-3x-10。

$\begin{aligned} Factor all denominators: & \frac{3}{x + 2} - \frac{4}{x - 5} = \frac{2}{(x + 2) (x - 5)} \\ Find the lowest common denominator: & LCD = (x + 2) (x - 5) \\ Multiply all terms in the equation by the LCD: \end{aligned}$

::系数所有分母 : 3x+2 - 4x - 5= 2x+2 (x-5) 确定最小公分母 : LCD= (x+2)(x- 5)

$\begin{aligned} (x + 2) (x - 5) \cdot \frac{3}{x + 2} - (x + 2) (x - 5) \cdot \frac{4}{x - 5} = (x + 2) (x - 5) \cdot \frac{2}{(x + 2) (x - 5)} \end{aligned}$

:x+2)(x-5)(x-5)(x-5)(x-5)(x+2)(x-5)(x-5)(x+2)(x-5)(x-2)(x-5)(x+2)(x+2)(x-2)(x-5))

$\begin{aligned} The equation simplifies to: & 3 (x - 5) - 4 (x + 2) = 2 \\ Simplify: & 3 x - 15 - 4 x - 8 = 2 \\ \underline{\underline{x = - 25}} \end{aligned}$

::等式简化为: 3(x- 5) - 4(x+2) = 2 简化: 3x- 15- 4x- 8= 2x_ 25__

Check: $\begin{aligned} \frac{3}{x + 2} - \frac{4}{x - 5} = \frac{2}{x^{2} - 3 x - 10} \Rightarrow \frac{3}{- 25 + 2} - \frac{4}{- 25 - 5} \overset{?}{=} \frac{2}{(- 25)^{2} - 3 (- 25) - 10} \Rightarrow .003 = .003 . \end{aligned}$ The answer checks out.
::检查:3x+2 - 4x - 5=2x2 - 3x - 103 - 25+2 - 4 - 25 - 5=? (2- 25)2 - 3(- 25) - 10N3=. 003。检查答案。

2. Solve $\begin{aligned} \frac{2 x}{2 x + 1} + \frac{x}{x + 4} = 1 \end{aligned}$ .
::2. 解决2x2x+1+xx+4=1。

$\begin{aligned} Find the lowest common denominator: & LCD = (2 x + 1) (x + 4) \\ Multiply all terms in the equation by the LCD: \end{aligned}$

::查找最小公分母 : LCD= (2x+1)(x+4) 用 LCD 查找方程式中的所有条件 :

$\begin{aligned} (2 x + 1) (x + 4) \cdot \frac{2 x}{2 x + 1} + (2 x + 1) (x + 4) \cdot \frac{x}{x + 4} = (2 x + 1) (x + 4) \end{aligned}$

:2x+1)(x+4)2x2xx+1+2(2x+1)(x+4)xxxx+4=(2xx+1)(x+4)xxx+4=(2x+1)(x+4))

$\begin{aligned} Cancel all common terms. & 2 x (x + 4) + x (2 x + 1) = (2 x + 1) (x + 4) \\ The simplified equation is: \\ Eliminate parentheses: & 2 x^{2} + 8 x + 2 x^{2} + x = 2 x^{2} + 9 x + 4 \\ Collect like terms: & 2 x^{2} = 4 \\ x^{2} = 2 \Rightarrow \underline{\underline{x = \pm \sqrt{2}}} \end{aligned}$

::取消所有共同的术语 2x( x+4) +x( 2x+1) =( 2x+1) (x+4) 简化方程式为: 取消括号 : 2x2+8x+2x2+2x2+x2+x=2x2+9x+4Collection like terms: 2x2=4x2=2\\\\%2___

Check: $\begin{aligned} \frac{2 x}{2 x + 1} + \frac{x}{x + 4} = \frac{2 \sqrt{2}}{2 \sqrt{2} + 1} + \frac{\sqrt{2}}{\sqrt{2} + 4} = 0.739 + 0.261 = 1. \end{aligned}$ The answer checks out.
::检查: 2x2x+1+xx+4=2222+1+22+4=0. 739+0. 261=1. 检查答案。

$\begin{aligned} \frac{2 x}{2 x + 1} + \frac{x}{x + 4} = \frac{2 (- \sqrt{2})}{2 (- \sqrt{2}) + 1} + \frac{- \sqrt{2}}{- \sqrt{2} + 4} = 1.547 - 0.547 = 1. \end{aligned}$ The answer checks out.
::2x2x+1+xx+4+4=2(-2)(-2)(-2)+12-2+4=1.547-0.547=1. 答案核对。

Examples
::实例

Solve the following rational equations :
::解决以下合理方程式:

Example 1
::例1

$\begin{aligned} \frac{3}{5} x + \frac{1}{2} = 4 \end{aligned}$
::35x+12=4

$\begin{aligned} Start with the original equation: & \frac{3}{5} x + \frac{1}{2} = 4 \\ Multiply by the LCD: & 10 \cdot (\frac{3}{5} x + \frac{1}{2}) = 10 \cdot 4 \\ Simplify: & 6 x + 5 = 40 \\ Isolate x first by subtracting 5 from each side: & 6 x + 5 - 5 = 40 - 5 \\ Simplify: & 6 x = 35 \\ Isolate x by dividing by 6: & \frac{6 x}{6} = \frac{35}{6} \\ Simplify: & x = 5 \frac{5}{6} \end{aligned}$

::以原始方程: 35x+12=4 开始, 以 LCD: 10( 35x+12) =10}4=10}4 简化: 6x+5=40 孤立 x 首先从每侧减去 5: 6x+5=5=40-5=40-5 简化: 6x=35Isolarate x 以 6: 6x6=356 简单化: x=556

Example 2
::例2

$\begin{aligned} \frac{x}{x - 1} + 1 = \frac{1}{x + 3} \end{aligned}$
::xx-1+1=1x+3

$\begin{aligned} Start with the original equation: & \frac{x}{x - 1} + 1 = \frac{1}{x + 3} . \\ Multiply by the LCD: & (x - 1) (x + 3) \cdot (\frac{x}{x - 1} + 1) = (x - 1) (x + 3) \cdot \frac{1}{2 x + 3} . \\ Simplify: & x (x + 3) + 1 (x - 1) (x + 3) = 1 (x - 1) \\ Distribute: & x^{2} + 3 x + x^{2} + 3 x - 1 x - 3 = x - 1 \\ Combine like terms: & 2 x^{2} + 5 x - 3 = x - 1 \\ Set one side equal to 0: & 2 x^{2} + 4 x + - 2 = 0 \end{aligned}$

::以原始方程开始 :xx-1+1+1=1x3. 以 LCD 开始 : (x-1)(x+3) = (x-1)(x-1-1) }(x-1-1+1) = (x-1) 12x+3. 简化 : x(x+3) +1(x-1)(x-1)(x+3) = 1(x-1) =1(x-1) 分发: x2+3x+x2+3x2+3x1x3=x-1Combine 类似条件: 2x2+5x3=x -1Set 一面等于 0: 2x2+4xx2=0

Now we have a quadratic equation . Since it's not factorable (check for yourself!), we must use the quadratic formula .
::现在我们有了四方方程式。因为这个方程式是不可计数的( 请自己检查! ) , 我们必须使用四方方程式。

$\begin{aligned} Start with the quadratic formula. x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Substitute in the appropriate values. x & = \frac{- 4 \pm \sqrt{4^{2} - 4 (2) (- 2)}}{2 (2)} \\ Simplify. x & = \frac{- 4 \pm \sqrt{32}}{4} = \frac{- 4 \pm 4 \sqrt{2}}{4} = - 1 \pm \sqrt{2} \end{aligned}$

::以二次方程式.xbb2-4ac2开始,以适当值作为替代品。x442-4(2)(2)-2(2)简化。x43244442412。

This means that $\begin{aligned} x = - 1 + \sqrt{2} \approx 0.4 \end{aligned}$ or $\begin{aligned} x = - 1 - \sqrt{2} \approx - 2.4 \end{aligned}$
::这意味着 x1+20.4 或 x1-22.4

Review
::回顾

Solve the following equations.
::解决以下方程式。

$\begin{aligned} \frac{2 x + 1}{4} = \frac{x - 3}{10} \end{aligned}$
::2x+14=x-310

$\begin{aligned} \frac{4 x}{x + 2} = \frac{5}{9} \end{aligned}$
::4xx+2=59

$\begin{aligned} \frac{5}{3 x - 4} = \frac{2}{x + 1} \end{aligned}$
::53x-4=2x+1

$\begin{aligned} \frac{7}{x + 3} = \frac{x + 1}{2 x - 3} \end{aligned}$
::7x+3=x+12x-3

$\begin{aligned} \frac{7 x}{x - 5} = \frac{x + 3}{x} \end{aligned}$
::7xx-5=x+3x

$\begin{aligned} \frac{2}{x + 3} - \frac{1}{x + 4} = 0 \end{aligned}$
::2x+3 - 1x+4=0

$\begin{aligned} \frac{3}{2 x - 1} + \frac{2}{x + 4} = 2 \end{aligned}$
::32 - 1+2x+4=2

$\begin{aligned} \frac{2 x}{x - 1} - \frac{x}{3 x + 4} = 3 \end{aligned}$
::2xx-1-x3x+4=3

$\begin{aligned} \frac{x + 1}{x - 1} + \frac{x - 4}{x + 4} = 3 \end{aligned}$
::x+1x-1+x-4x+4=3 x1x-1+x-4x+4=3

$\begin{aligned} \frac{x}{x - 2} + \frac{x}{x + 3} = \frac{1}{x^{2} + x - 6} \end{aligned}$
::xx-2+xxx+3=1x2+x-6

$\begin{aligned} \frac{2}{x^{2} + 4 x + 3} = 2 + \frac{x - 2}{x + 3} \end{aligned}$
::2x2+4x+3=2+x-2x+3

$\begin{aligned} \frac{1}{x + 5} - \frac{1}{x - 5} = \frac{1 - x}{x + 5} \end{aligned}$
::1x+5-1x-5=1-xxx+5

$\begin{aligned} \frac{x}{x^{2} - 36} + \frac{1}{x - 6} = \frac{1}{x + 6} \end{aligned}$
::x2-36+1x-6=1x+6

$\begin{aligned} \frac{2 x}{3 x + 3} - \frac{1}{4 x + 4} = \frac{2}{x + 1} \end{aligned}$
::2x3x+3-14x+4=2x+1

$\begin{aligned} \frac{- x}{x - 2} + \frac{3 x - 1}{x + 4} = \frac{1}{x^{2} + 2 x - 8} \end{aligned}$
::-xx-2+3x-1x+4=1x2+2x-8

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。

版权所有 © 2024 由瓜瓜科技thewaytolearn.cn设计。保留所有权利。