Course: 10.11 部分分数分解

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Introduction
::导言

Suppose you are painting a very large mural on the side of a building. You need to calculate the area under a curve in the mural, so you know how much paint to buy. In calculus, you'll learn how to use a method called "integration" to determine the area under a curve. If the curve is very complex, breaking it into parts will simplify the integration process. Partial fractions are used to break rational equations into parts.
::假设您正在建筑的侧面画一个非常大的壁画。您需要在壁画的曲线下计算区域, 以便您知道要购买多少油漆。在微积分中, 您将学会如何使用一个叫做“ 集成” 的方法来确定曲线下的区域。如果曲线非常复杂, 将其分割成一些部分将简化集成过程。部分分数用于将理性方程式分解成几个部分。

When given a rational expression such as $\begin{aligned} \frac{4 x - 9}{x^{2} - 3 x} \end{aligned}$ , we can break it into the sum of two simpler fractions . The challenging part is trying to get from the initial rational expression to the simpler fractions. You may know how to add fractions and go from two or more separate fractions to a single fraction , but how do you reverse this process?
::当给定4x- 9x2-3x 等合理表达式时, 我们可以把它折成两个简单分数的总和。挑战部分是试图从最初的理性表达式到简单的分数。您可能知道如何添加分数, 从两个或两个以上的分数转到一个分数, 但是您如何扭转这个过程 ?

Partial Fractions
::部分分数

Partial fraction decomposition , also referred to as partial fraction expansion, is a procedure that reverses adding fractions with unlike denominators. The most challenging part is coming up with the denominators of each individual partial fraction. See if you can spot the pattern .
::部分分解(也称为部分分解扩展)是一种以不同分母反向增加分数的程序。最具挑战性的部分是每个部分分母的分母。看看您能否辨别出模式。

$\begin{aligned} \frac{6 x - 1}{x^{2} (x - 1) (x^{2} + 2)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 1} + \frac{D x + E}{x^{2} + 2} \end{aligned}$

::6-1x2x2(x-2)-1(x2+2)=Ax+Bx2+Cx-1+Dx+Ex2+2

Notice how, in this example, each individual factor was represented. When performing partial fraction decomposition, linear factors that are raised to a power greater than one must have each successive power included as a separate denominator. These factors are called repeated factors. When the denominator of the rational expression has a repeated factor, a factor for each power needs to be included in the partial fraction expansion. For example,
::在此示例中, 注意每个单个系数的表示方式。当进行部分分数分解时, 升至一个以上功率的线性系数必须包含每个连续功率作为单独的分母。这些因素被称为重复因素。当理性表达的分母有一个重复系数时, 每项功率的一个系数必须包含在部分分数扩展中。例如,

$\begin{aligned} \frac{1}{(x - 2)^{3}} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^{2}} + \frac{C}{(x - 2)^{3}} . \end{aligned}$

::1(x-2)3=A(x-2)+B(x-2)2+C(x-2)3。

Also from the initial example, quadratic terms that do not factor to be linear terms are included with a numerator that is a linear function of $\begin{aligned} x \end{aligned}$ . For example,
::另外,从最初的示例中,不包括线性术语因素的二次词组也包含一个X的线性函数的分子。例如,

$\begin{aligned} \frac{1}{x (x^{2} + 6)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 6} . \end{aligned}$

::1(x2+6)=Ax+Bx+Cx2+6。

In general, a rational function $\begin{aligned} \frac{P (x)}{Q (x)} \end{aligned}$ can be rewritten using what is known as partial fraction decomposition. This procedure often allows integration to be performed on each term separately by inspection. For each factor of $\begin{aligned} Q (x), \end{aligned}$ the form $\begin{aligned} {(a x + b)}^{m} \end{aligned}$ introduces terms
::一般说来,理性函数P(x)Q(x) (x) 可以用所谓的部分分分解法重写。该程序通常允许通过检查对每个术语分别进行整合。对于Q(x) 的每一个系数,形式(ax+b)m 引入术语。

$\begin{aligned} \frac{A_{1}}{a x + b} + \frac{A_{2}}{{(a x + b)}^{2}} + . . . + \frac{A_{m}}{{(a x + b)}^{m}} . \end{aligned}$

::A1ax+b+A2(ax+b)2+...+AAM(ax+b)m.

Then write
::然后写

$\begin{aligned} \frac{P (x)}{Q (x)} = \frac{A_{1}}{a x + b} + . . . + \frac{A_{2} x + B_{2}}{{a x}^{2} + b x + c} + . . . \end{aligned}$

::P(x)Q(x) = A1ax+b+...+A2x+B2ax2+bx+c+...

and solve for all $\begin{aligned} A_{i} \end{aligned}$ and $\begin{aligned} B_{i} \end{aligned}$ .
::并且为爱爱和白爱而破晓。

The following video demonstrates how to perform partial fraction decomposition:
::以下视频展示如何进行部分分解:

Examples
::实例

Example 1
::例1

Add the partial fractions to show both sides are equal.
::添加部分分数以显示两侧相等。

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

::2x+4(x- 1)(x+3)=1x+1+1+1x+3

Solution:
::解决方案 :

$\begin{aligned} \frac{1}{x + 1} + \frac{1}{x + 3} = \frac{x + 3}{(x + 1) (x + 3)} + \frac{x + 1}{(x + 1) (x + 3)} = \frac{2 x + 4}{(x + 1) (x + 3)} \end{aligned}$

::1x+1+1+1+3=x3+3(x+1)(x+3)+x+1(x+1)(x+3)=2x+4(x+1)(x+3)

Example 2
::例2

Recall the problem from the Introduction: W hat is the partial fraction decomposition of the given rational expression?
::回顾导言中的问题:给定的理性表达的局部分部分分解是什么?

$\begin{aligned} \frac{4 x - 9}{x^{2} - 3 x} \end{aligned}$
Solution:
::4 - 9x2 - 3 解析器 :

To decompose the rational expression into the sum of two simpler fractions, you need to use partial fraction decomposition.
::要将理性表达方式分解为两个简单分数的总和, 您需要使用部分分数分解法。

$\begin{aligned} \frac{4 x - 9}{x^{2} - 3 x} & = \frac{A}{x} + \frac{B}{x - 3} \\ 4 x - 9 & = A (x - 3) + B x \end{aligned}$

::4-9x2-3x=Ax+Bx-34x-9=A(x-3)+Bx

When $\begin{aligned} x = 3 \end{aligned}$ , the factor $\begin{aligned} (x - 3) \end{aligned}$ equals 0, and we can solve for $\begin{aligned} B \end{aligned}$ .
::当 x= 3 时, 系数( x-3) 等于 0, 我们可以解决 B 。

$\begin{aligned} 4 \cdot 3 - 9 & = A (3 - 3) + B \cdot 3 \\ 3 & = 3 B \\ B & = 1 \end{aligned}$

When $\begin{aligned} x = 0 \end{aligned}$ , we can solve for $\begin{aligned} A \end{aligned}$ .
::43-9=A(3-3)+B33=3BBB=1 x=0,我们可以解答 A。

$\begin{aligned} 4 \cdot 0 - 9 & = A (0 - 3) + B \cdot 0 \\ - 9 & = - 3 A \\ A & = 3 \end{aligned}$

::40-9=A(0-3)+B0-93A=3

Solving this system yields $\begin{aligned} A = 3 and B = 1. \end{aligned}$ Therefore ,
::解决这一系统产生A=3和B=1。因此,

$\begin{aligned} \frac{4 x - 9}{x^{2} - 3 x} = \frac{3}{x} + \frac{1}{x - 3} . \end{aligned}$
Example 3
::4-9x2-3x=3x+1x-3. 例3

Use partial fractions to decompose the rational expression.
::使用部分分数分解正态表达式。

$\begin{aligned} \frac{7 x^{2} + x + 6}{x^{3} + 3 x} \end{aligned}$

::7x2+x+6x3+3x

Solution:
::解决方案 :

First, factor the denominator and identify the denominators of the partial fractions.
::首先,乘数分母并确定部分分数的分母。

$\begin{aligned} \frac{7 x^{2} + x + 6}{x (x^{2} + 3)} = \frac{A}{x} + \frac{B x + C}{x^{2} + 3} \end{aligned}$

::7x2+x+6x(x2+3)=Ax+Bx+Cx2+3

When the fractions are eliminated by multiplying through by the LCD, the equation becomes
::当分数通过LCD的乘法消除分数时,方程式就变成

$\begin{aligned} 7 x^{2} + x + 6 & = A (x^{2} + 3) + x (B x + C) \\ 7 x^{2} + x + 6 & = A x^{2} + 3 A + B x^{2} + C x . \end{aligned}$

::7x2+x+6=A(x2+3)+x(Bx+C)7x2+6=Ax2+3A+Bx2+Cx。

Notice that the squared term, linear term, and constant term form a system of three equations with three variables.
::注意正方形术语、线性术语和恒定术语形成由三个方程式组成的系统,有三个变量。

$\begin{aligned} A + B & = 7 \\ C & = 1 \\ 3 A & = 6 \end{aligned}$

::A+B=7C=13A=6 A+B=7C=13A=6

In this case, it is easy to see that $\begin{aligned} A = 2, B = 5, C = 1 \end{aligned}$ . Often, the resulting system of equations is more complex and would benefit from your knowledge of solving systems using matrices.
::在这种情况下,很容易看到A=2,B=5,C=1。通常,由此产生的方程系统更为复杂,并会受益于你对使用矩阵解决系统的知识。

$\begin{aligned} \frac{7 x^{2} + x + 6}{x (x^{2} + 3)} = \frac{2}{x} + \frac{5 x + 1}{x^{2} + 3} \end{aligned}$

::7x2+x+6x(x2+3)=2x+5x+1x2+3

Example 4
::例4

Decompose the rational expression.
::使理性表达方式分解。

$\begin{aligned} \frac{5 x^{4} - 3 x^{3} - x^{2} + 4 x - 1}{(x - 1)^{3} x^{2}} \end{aligned}$

::5x4-3x3-x2+4x-1(x-1-1)3x2

Solution:
::解决方案 :

First, identify the denominators of the partial fractions. Note that there are repeated factors in the denominator of the given rational expression.
::首先,确定部分分数的分母。请注意,在给定的理性表达的分母中存在反复发生的因素。

$\begin{aligned} \frac{5 x^{4} - 3 x^{3} - x^{2} + 4 x - 1}{(x - 1)^{3} x^{2}} = \frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{(x - 1)^{3}} + \frac{D}{x} + \frac{E}{x^{2}} \end{aligned}$

::5x4-3x3-x2+4x-1(x-1)3x2=Ax-1+B(x-1-1)2+C(x-1)3+Dx+Ex2

When the entire fraction is multiplied through by $\begin{aligned} (x - 1)^{3} x^{2}, \end{aligned}$ the equation results to
::当整个分数乘以 (x- 1) 3x2 时, 方程结果为

$\begin{aligned} 5 x^{4} - 3 x^{3} - x^{2} + 4 x - 1 \\ = A (x - 1)^{2} x^{2} + B (x - 1) x^{2} + C x^{2} + D (x - 1)^{3} x + E (x - 1)^{3} . \end{aligned}$

::5x4-3x3-x2+4x-1=A(x-1)2x2+B(x-1)x2+Cx2+D(x-1)3x+E(x-1)3。

Multiplication of each term can be done separately to be extra careful.
::每个术语的乘法可以分开进行,以格外谨慎。

$\begin{aligned} A x^{4} - 2 A x^{3} + A x^{2} \\ B x^{3} - B x^{2} \\ C x^{2} \\ D x^{4} - 3 D x^{3} + 3 D x^{2} - D x \\ E x^{3} - 3 E x^{2} + 3 E x - E \end{aligned}$

::Ax4-2Ax3+Ax2Bx3-Bx2Bx3-Bx2C2x2Dx4-D3x3+3Dx2-Dx3-3Dx2-DxEx3-3Ex3-3Ex2+3-E

Group terms with the same power of $\begin{aligned} x, \end{aligned}$ and set equal to the corresponding term.
::x具有相同功率的组名, 并设定等于相应术语的组名。

$\begin{aligned} 5 x^{4} & = A x^{4} + D x^{4} \\ - 3 x^{3} & = - 2 A x^{3} + B x^{3} - 3 D x^{3} + E x^{3} \\ - x^{2} & = A x^{2} - B x^{2} + C x^{2} + 3 D x^{2} - 3 E x^{2} \\ 4 x & = - D x + 3 E x \\ - 1 & = - E \end{aligned}$

::5x4=Ax4+Dx4-3x3}2Ax3+Bx3-3Dx3+Ex3-x2=Ax2-Bx2+Cx2+3Dx2+3Dx2-3Ex24x}Dx+3Ex1__E

From these 5 equations, every $\begin{aligned} x \end{aligned}$ can be divided out. Assume that $\begin{aligned} x \neq 0 \end{aligned}$ because if it were, then the original expression would be undefined .
::从这5个方程式中, 每个 x 都可以被分割。假设 x+++0 因为如果是的话, 那么原来的表达式就会没有定义。

$\begin{aligned} 5 & = A + D \\ - 3 & = - 2 A + B - 3 D + E \\ - 1 & = A - B + C + 3 D - 3 E \\ 4 & = - D + 3 E \\ - 1 & = - E \end{aligned}$

::5=A+D-3=D-32A+B-3D+E-1=A-B+C+3D-3E4+3E-1E

This is a system of equations of 5 variables and 5 equations. Some of the equations can be solved using logic and substitution, like $\begin{aligned} E = 1, D = - 1, A = 6 \end{aligned}$ . You can use any method involving or matrices. In this case, it is easiest to substitute known values into equations with one unknown value to get more known values and repeat.
::这是一个由 5 个变量和 5 个方程式组成的方程式系统。有些方程式可以通过逻辑和替代( 如 E=1, D1, A=6) 解析。您可以使用任何涉及或矩阵的方法。在这种情况下, 最容易将已知值替换成一个未知值的方程式, 以获得更多已知值和重复。

$\begin{aligned} B & = 5 \\ C & = 4 \\ \frac{5 x^{4} - 3 x^{3} - x^{2} + 4 x - 1}{(x - 1)^{3} x^{2}} & = \frac{6}{x - 1} + \frac{5}{(x - 1)^{2}} + \frac{4}{(x - 1)^{3}} + \frac{- 1}{x} + \frac{1}{x^{2}} \end{aligned}$

::B=5C=45C=45C4-3x4-3x3-x2+4x1-1(x-1-3x2=6x1+5(x-1-1)2+4(x-1)3x1x1x1x1x2

Example 5
::例5

Use matrices to complete the partial fraction decomposition of the rational expression.
::使用矩阵来完成合理表达式的局部分解。

$\begin{aligned} \frac{2 x + 4}{(x - 1) (x + 3)} \end{aligned}$

::2x+4(x-1)(x+3)

Solution:
::解决方案 :

$\begin{aligned} \frac{2 x + 4}{(x - 1) (x + 3)} & = \frac{A}{x + 1} + \frac{B}{x + 3} \\ 2 x + 4 & = A x + 3 A + B x + B \end{aligned}$

::2x+4(x-1)(x+3)=Ax+1+Bx+Bx+32x+4=Ax+3A+Bx+B

$\begin{aligned} 2 & = A + B \\ 4 & = 3 A + B \end{aligned}$

::2=A+B4=3A+B

$\begin{aligned} [\begin{matrix} \begin{matrix} 1 & 1 \\ 3 & 1 \end{matrix} | \begin{matrix} 2 \\ 4 \end{matrix} \end{matrix}] \begin{matrix} \to \\ - 3 \cdot R_{1} & \to \end{matrix} & [\begin{matrix} \begin{matrix} 1 & 1 \\ 0 & - 2 \end{matrix} | \begin{matrix} 2 \\ - 2 \end{matrix} \end{matrix}] \begin{matrix} \to \\ \div - 2 & \to \end{matrix} [\begin{matrix} \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}] \begin{matrix} - R_{2} & \to \\ \to \end{matrix} [\begin{matrix} \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix}] \end{aligned}$

::[1131]_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} A = 1, & B = 1 \end{aligned}$

::A=1, B=1

$\begin{aligned} \frac{2 x + 4}{(x - 1) (x + 3)} & = \frac{1}{x + 1} + \frac{1}{x + 3} \end{aligned}$

::2x+4(x- 1)(x+3)=1x+1+1+1x+3

Example 6
::例6

Use matrices to help you decompose the rational expression.
::使用矩阵帮助您分解合理表达式。

$\begin{aligned} \frac{5 x - 2}{(2 x - 1) (3 x + 4)} \end{aligned}$

::5x-2(2-2x-1)(3x+4)

Solution:
::解决方案 :

$\begin{aligned} \frac{5 x - 2}{(2 x - 1) (3 x + 4)} & = \frac{A}{2 x - 1} + \frac{B}{3 x + 4} \\ 5 x - 2 & = A (3 x + 4) + B (2 x - 1) \\ 5 x - 2 & = 3 A x + 4 A + 2 B x - B \\ 5 & = 3 A + 2 B \\ - 2 & = 4 A - B \end{aligned}$

$\begin{aligned} [\begin{matrix} \begin{matrix} 3 & 2 \\ 4 & - 1 \end{matrix} | \begin{matrix} 5 \\ - 2 \end{matrix} \end{matrix}] & \begin{matrix} \cdot 4 & \to \\ \cdot 3 & \to \end{matrix} [\begin{matrix} \begin{matrix} 12 & 8 \\ 12 & - 3 \end{matrix} | \begin{matrix} 20 \\ - 6 \end{matrix} \end{matrix}] \begin{matrix} \to \\ - R_{1} & \to \end{matrix} [\begin{matrix} \begin{matrix} 12 & 8 \\ 0 & - 11 \end{matrix} | \begin{matrix} 20 \\ - 26 \end{matrix} \end{matrix}] \\ \begin{matrix} \to & \cdot 11 & \to \\ \to & \cdot 8 & \to \end{matrix} & [\begin{matrix} \begin{matrix} 132 & 88 \\ 0 & - 88 \end{matrix} | \begin{matrix} 220 \\ - 208 \end{matrix} \end{matrix}] \begin{matrix} + R_{2} & \to \\ \to \end{matrix} [\begin{matrix} \begin{matrix} 132 & 0 \\ 0 & - 88 \end{matrix} | \begin{matrix} 12 \\ - 208 \end{matrix} \end{matrix}] \\ \begin{matrix} \to & \div 132 & \to \\ \to & \div - 88 & \to \end{matrix} [\begin{matrix} \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | \begin{matrix} \frac{1}{11} \\ \frac{26}{11} \end{matrix} \end{matrix}] \end{aligned}$

::5x-2(2x-2x-1)(3x+4)=A2x-1-1(3x+4)=A2x-1+B3x-4)=A2x-1-1(3x+B3x-2x-2x-2x-1)=A2x-2x-2x-2x-2x-1(2x-2x-2x-1)(3x+4)=A2x-1-1(3x+4)=A2x-1(3x+B2x-1)=A2=3A+4A+2Bx-B5=3A-2B-2B-2=3A-2B-2B-2=4A-B[324-1-1-5-5-5-2,5-2)44443}[1281212-3-3}[206}}R1}[1280-111__2026}*11}[13880-88_20_20-88)。

$\begin{aligned} A = \frac{1}{11}, & B = \frac{26}{11} \\ \frac{5 x - 2}{(2 x - 1) (3 x + 4)} & = \frac{\frac{1}{11}}{2 x - 1} + \frac{\frac{26}{11}}{3 x + 4} \end{aligned}$

::A=111, B=26115x-2-2-2x-1(3x+4) =1112x-1+26113x+4

Example 7
::例7

Add the partial fractions to check Example 6.
::添加部分分数以检查例6。

$\begin{aligned} \frac{5 x - 2}{(2 x - 1) (3 x + 4)} = \frac{\frac{1}{11}}{2 x - 1} + \frac{\frac{26}{11}}{3 x + 4} \end{aligned}$

::5x-2-2x-1(3x+4)=1112x-1+26113x+4

Solution:

$\begin{aligned} \frac{5 x - 2}{(2 x - 1) (3 x + 4)} & = \frac{\frac{1}{11}}{2 x - 1} + \frac{\frac{26}{11}}{3 x + 4} \\ 5 x - 2 & = \frac{1}{11} (3 x + 4) + \frac{26}{11} (2 x - 1) \\ 55 x - 22 & = 3 x + 4 + 26 (2 x - 1) \\ 55 x - 22 & = 3 x + 4 + 52 x - 26 \\ 55 x - 22 & = 55 x - 22 \end{aligned}$

::溶解度: 5x-2-2x-1(3x+4) = 1112x-1+26113x+45x-2= 111(3x+4) +2611(2x-1) 55x-22= 3x-4+26(2x-1) 55x-22= 3x-4+4(2x-1) 55x-22= 3x+4+52x-225x-25x=55x-22

Summary
::摘要

Partial fraction decomposition is a procedure that undoes the operation of adding fractions with unlike denominators.
::部分碎片分解是一种程序,可以取消与分母不同的添加分数的操作。

Partial fraction decomposition separates a rational expression into the sum of rational expressions with unlike denominators.
::部分分解分解将合理表达式分离成与分母不同的合理表达式和合理表达式之和。

When performing partial fraction decomposition, repeated factors , linear factors that are raised to a power greater than one, must have each successive power included as a separate denominator.
::当进行部分分解时,重复性因素,即升至大于一个功率的线性因素,必须把每一连续的功率作为一个单独的分母列入。

When performing partial fraction decomposition, quadratic terms that do not factor to be linear terms are included with a numerator that is a linear function of $\begin{aligned} x \end{aligned}$ .
::当进行部分分解分解时,不包括线性术语因素的二次曲线术语与X的线性函数分子一起列入。

Review
::回顾

Decompose the rational expressions below. Practice using matrices with at least one of the problems.
::将合理表达方式分解如下。使用矩阵的做法至少有一个问题。

1. $\begin{aligned} \frac{3 x - 4}{(x - 1) (x + 4)} \end{aligned}$
::1. 3x-4(x-1)(x+4)

2. $\begin{aligned} \frac{2 x + 1}{x^{2} (x - 3)} \end{aligned}$
::2. 2x+1x2(x-3)

3. $\begin{aligned} \frac{x + 1}{x (x - 5)} \end{aligned}$
::3. x+1x(x-5)

4. $\begin{aligned} \frac{x^{2} + 3 x + 1}{x (x - 3) (x + 6)} \end{aligned}$
::4. x2+3x+1x(x-3)(x+6)

5. $\begin{aligned} \frac{3 x^{2} + 2 x - 1}{x^{2} (x + 2)} \end{aligned}$
::5. 3x2+2x-1x2(x+2)

6. $\begin{aligned} \frac{x^{2} + 1}{x (x - 1) (x + 1)} \end{aligned}$
::6. x2+1x(x-1)(x+1)

7. $\begin{aligned} \frac{4 x^{2} - 9}{x^{2} (x - 4)} \end{aligned}$
::7. 4x2-9x2(x-4)

8. $\begin{aligned} \frac{2 x - 4}{(x + 7) (x - 3)} \end{aligned}$
::8. 2x-4(x+7)(x-3)

9. $\begin{aligned} \frac{3 x - 4}{x^{2} (x^{2} + 1)} \end{aligned}$
::9. 3x-4x2(x2+1)

10. $\begin{aligned} \frac{2 x + 5}{(x - 3) (x^{2} + 4)} \end{aligned}$
::10. 2x+5(x-3)(x2+4)

11. $\begin{aligned} \frac{3 x^{2} + 2 x - 5}{x^{2} (x - 3) (x^{2} + 1)} \end{aligned}$
::11. 3x2+2x-5x2(x-3)(x2+1)

12. Confirm your answer to Number 1 by adding the partial fractions.
::12. 通过添加部分分数确认您对第1号的答复。

13. Confirm your answer to Number 3 by adding the partial fractions.
::13. 通过添加部分分数确认你对第3号的答复。

14. Confirm your answer to Number 6 by adding the partial fractions.
::14. 通过添加部分分数确认你对第6号的答复。

15. Confirm your answer to Number 9 by adding the partial fractions.
::15. 通过添加部分分数确认你对第9号的答复。

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

General

分数

Introduction ::导言

Partial Fractions ::部分分数

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Partial Fractions
::部分分数

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)