课程： 8.10 部分分数 | The Way To Learn

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

部分分数

When given a rational expression like $\begin{align*}\frac{4x-9}{x^2-3x}\end{align*}$ it is very helpful in calculus to be able to write it as the sum of two simpler fractions like $\begin{align*}\frac{3}{x}+\frac{1}{x-3}\end{align*}$ . The challenging part is trying to get from the initial rational expression to the simpler fractions.
::当给定了 4x- 9x2- 3x 等合理表达式时, 可以用微积分写成两个简单分数的总和( 如 3x+1x-3) 来帮助计算。挑战部分是试图从最初的合理表达式到简单的分数。

You may know how to add fractions and go from two or more separate fractions to a single fraction, but how do you go the other way around?
::你也许知道如何添加分数, 从两个或两个以上的分数移到一个分数, 但你如何绕到另一个方向呢?

Partial Fraction Decomposition
::分数分数分数分数分数分数分数

Partial fraction decomposition 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.
::部分分解是一种程序,它会以不同的分母来反向增加分分母。最具挑战性的部分是每个部分的分母。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}

$\begin{align*}\frac{6x-1}{x^2 (x-1)(x^2+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{Dx+E}{x^2+2}\end{align*}$
::6-1x2x2(x-2)-1(x2+2)=Ax+Bx2+Cx-1+Dx+Ex2+2

In this example each individual factor of the denominator must be represented. Linear factors that are raised to a power greater than one must have each successive power included as a separate denominator. Quadratic terms that do not factor to be linear terms are included with a numerator that is a linear function of $\begin{align*}x\end{align*}$ . Take a look at the examples to see partial fraction decomposition put into practice.
::在此示例中,分母的每个单个系数都必须被代表。升至大于一个功率的线性系数必须将每一相继功率单独列为分母。不包括线性术语的二次曲线术语包含一个X线性函数的分子。查看这些示例,看将部分分分解纳入实践。

Examples
::实例

Example 1
::例1

Earlier, you were asked how to go from one fraction to multiple simpler fractions. To decompose the rational expression into the sum of two simpler fractions you need to use partial fraction decomposition.
::早些时候, 您被问及如何从一个分数转换到多个简单分数。要将理性表达方式分解成两个简单分数的总和, 您需要使用部分分数分解。

\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 \\ 4 x - 9 & = A x - 3 A + B x \end{aligned}

$\begin{align*}\frac{4x-9}{x^2-3x} &= \frac{A}{x}+\frac{B}{x-3}\\ 4x-9 &= A(x-3)+Bx\\ 4x-9 &= Ax-3A+Bx\end{align*}$
::4-9x2-3x=Ax+Bx-34x-9=A(x-3)+Bx4x-9=Ax-3A+Bx

Notice that the constant term -9 must be equal to the constant term $\begin{align*}-3A\end{align*}$ and that the terms with $\begin{align*}x\end{align*}$ must be equal as well.
::请注意,常数-9必须等于常数-3A,与x的条件也必须相等。

\begin{aligned} - 9 & = - 3 A \\ 4 & = A + B \end{aligned}

$\begin{align*}-9 &= -3A\\ 4 &= A+B\end{align*}$
::-9*3A4=A+B

Solving this system yields:
::解决这个系统的结果:

$\begin{align*}A=3, \quad B=1\end{align*}$
::A=3,B=1,A=3,B=1

Therefore,
::因此,

$\begin{align*}\frac{4x-9}{x^2-3x}=\frac{3}{x}+\frac{1}{x-3}\end{align*}$
::4-9x2-3x=3x+1x-3

Example 2
::例2

Use partial fractions to decompose the following rational expression.
::使用部分分数分解以下合理表达式的分解。

$\begin{align*}\frac{7x^2+x+6}{x^3+3x}\end{align*}$
::7x2+x+6x3+3x

First factor the denominator and identify the denominators of the partial fractions.
::第一个因素是分母,并确定部分分母的分母。

$\begin{align*}\frac{7x^2+x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}\end{align*}$
::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的乘法消除时,方程式变成:

\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}

$\begin{align*}7x^2+x+6 &= A(x^2+3)+x(Bx+C)\\ 7x^2+x+6 &= Ax^2+3A+Bx^2+Cx\end{align*}$
::7x2+x+6=A(x2+3)+x(Bx+C)7x2+6=Ax2+3A+Bx2+Cx

Notice 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组成的系统,有三个变量。

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

$\begin{align*}A+B &= 7\\ C &= 1\\ 3A &= 6\end{align*}$
::A+B=7C=13A=6 A+B=7C=13A=6

In this case it is easy to see that $\begin{align*}A=2, B=5, C=1\end{align*}$ . 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。

$\begin{align*}\frac{7x^2+x+6}{x(x^2+3)}=\frac{2}{x}+\frac{5x+1}{x^2+3}\end{align*}$
::7x2+x+6x(x2+3)=2x+5x+1x2+3

Example 3
::例3

Decompose the following rational expression.
::使以下合理表达方式分解。

$\begin{align*}\frac{5x^4-3x^3-x^2+4x-1}{(x-1)^3x^2}\end{align*}$
::5x4-3x3-x2+4x-1(x-1-1)3x2

First identify the denominators of the partial fractions.
::首先确定部分分数的分母。

\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}

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

$\begin{align*}&5x^4-3x^3-x^2+4x-1\\&=A(x-1)^2x^2+B(x-1)x^2+Cx^2+D(x-1)^3x+E(x-1)^3\end{align*}$
::5x4-3x3-x2+4x-1=A(x-1)2x2+B(x-1)2x2+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}

$\begin{align*}& Ax^4-2Ax^3+Ax^2\\ & Bx^3-Bx^2\\ & Cx^2\\ & Dx^4-3Dx^3+3Dx^2-Dx\\ & Ex^3-3Ex^2+3Ex-E\end{align*}$
::Ax4-2Ax3+Ax2Bx3-Bx2Bx3-Bx2C2x2Dx4-D3x3+3Dx2-Dx3-3Dx2-DxEx3-3Ex3-3Ex2+3-E

Group terms with the same power of $\begin{align*}x\end{align*}$ and set equal to the corresponding term.
::x具有相同功率的组词组,与相应术语相同。

\begin{aligned} 5 x^{4} & = A x^{4} + D x^{4} \\ - 3 x^{3} & = - 2 A x^{3} + B x^{3} - 3 D^{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}

$\begin{align*}5x^4 &= Ax^4+Dx^4\\ -3x^3 &= -2Ax^3+Bx^3-3D^3+Ex^3\\ -x^2 &= Ax^2-Bx^2+Cx^2+3Dx^2-3Ex^2\\ 4x &= -Dx+3Ex\\ -1 &= -E\end{align*}$
::5x4 = Ax4+Dx4-3x32Ax3+Bx3-3D3+Ex3-x2=Ax2-Bx2+Cx2+3Dx2+3Dx2_3Ex24x}Dx+3Ex1*E

From these 5 equations, every $\begin{align*}x\end{align*}$ can be divided out. Assume that $\begin{align*}x \neq 0\end{align*}$ 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}

$\begin{align*}5 &= A+D\\ -3 &= -2A+B-3D+E\\ -1 &= A-B+C+3D-3E\\ 4 &= -D+3E\\ -1 &= E\end{align*}$
::5=A+D-3=D-32A+B-3D+E-1=A-B+C+3D-3E4+D-3E-1=E

This is a system of equations of five variables and 5 equations. Some of the equations can be solved using logic and substitution like $\begin{align*}E=-1, \ D=-7, \ A=12\end{align*}$ . 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.
::这是一个由五个变量和五个方程式组成的方程式系统。某些方程式可以通过逻辑和替代( 如 E1, D7, A=12) 解答。您可以使用任何涉及或矩阵的方法或矩阵。在此情况下, 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值。

\begin{aligned} B & = 1 \\ C & = 6 \\ \frac{5 x^{4} - 3 x^{3} - x^{2} + 4 x - 1}{(x - 1)^{3} x^{2}} & = \frac{12}{x - 1} + \frac{1}{(x - 1)^{2}} + \frac{6}{(x - 1)^{3}} + \frac{- 7}{x} + \frac{- 1}{x^{2}} \end{aligned}

$\begin{align*}B &= 1\\ C &= 6\\ \frac{5x^4-3x^3-x^2+4x-1}{(x-1)^3x^2} &= \frac{12}{x-1}+\frac{1}{(x-1)^2}+\frac{6}{(x-1)^3}+\frac{-7}{x}+\frac{-1}{x^2}\end{align*}$
::B=1C=65x4-3x3-x2+4x1(x-1-13x2=12x1+1(x-1-1)2+6(x-1)3_7x1x1x2

Example 4
::例4

Use matrices to complete the partial fraction decomposition of the following rational expression and confirm the solution.
::使用矩阵来完成以下合理表达式的局部分解,并确认解决方案。

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

$\begin{align*}\frac{2x+4}{(x+1)(x+3)}\end{align*}$
::2x+4( x+1)( x+3)

\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}

$\begin{align*}\frac{2x+4}{(x+1)(x+3)} &= \frac{A}{x+1}+\frac{B}{x+3}\\ 2x+4 &= Ax+3A+Bx+B\\ \end{align*}$
::2x+4(x+1)(x+3)=Ax+1+Bx+32x+4=Ax+3A+Bx+B

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

$\begin{align*}2 &= A+B\\ 4 &= 3A+B\\ \end{align*}$
::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 \\ \to & - 3 \cdot I & \to \end{matrix} & [\begin{matrix} \begin{matrix} 1 & 1 \\ 0 & - 2 \end{matrix} | \begin{matrix} 2 \\ - 2 \end{matrix} \end{matrix}] \begin{matrix} \to \\ \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} \to & - I I & \to \\ \to \end{matrix} [\begin{matrix} \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} | \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix}] \end{aligned}

$\begin{align*}\begin{bmatrix} \begin{matrix} 1 & 1\\ 3 & 1 \end{matrix} \left| \begin{matrix} 2\\ 4 \end{matrix}\right. \end{bmatrix} \begin{matrix} & \rightarrow &\\ \rightarrow & -3 \cdot I & \rightarrow \end{matrix} & \begin{bmatrix} \begin{matrix} 1 & 1\\ 0 & -2 \end{matrix} \left| \begin{matrix} 2\\ -2 \end{matrix}\right. \end{bmatrix} \begin{matrix} & \rightarrow &\\ \rightarrow & \div -2 & \rightarrow \end{matrix} \begin{bmatrix} \begin{matrix} 1 & 1\\ 0 & 1 \end{matrix} \left| \begin{matrix} 2\\ 1 \end{matrix}\right. \end{bmatrix} \begin{matrix} \rightarrow & -II & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix} \left| \begin{matrix} 1\\ 1 \end{matrix}\right. \end{bmatrix}\\\end{align*}$
::[1111] [1111] [1111] [1111] [1111] [1111] [1111] [1111]

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

$\begin{align*}A=1, \ & B=1\\\end{align*}$
::A=1, B=1

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

$\begin{align*}\frac{2x+4}{(x+1)(x+3)} &= \frac{1}{x+1}+\frac{1}{x+3}\end{align*}$
::2x+4( x+1)( x+3)=1x+1+1x+3

To confirm this answer, add the fractions.
::为了确认这一答复,加上分数。

$\begin{align*}\frac{1}{x+1}+\frac{1}{x+3}=\frac{x+3}{(x+1)(x+3)}+\frac{x+1}{(x+1)(x+3)}=\frac{2x+4}{(x+1)(x+3)}\end{align*}$
::1x+1+1+1+3=x3+3(x+1)(x+3)+x+1(x+1)(x+3)=2x+4(x+1)(x+3)

Example 5
::例5

Use matrices to help you decompose the following rational expression. Confirm the solution by adding the partial fractions.
::使用矩阵帮助您分解以下合理表达式。通过添加部分分数来确认解决方案。

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

\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 \\ [\begin{matrix} \begin{matrix} 3 & 2 \\ 4 & - 1 \end{matrix} | \begin{matrix} 5 \\ - 2 \end{matrix} \end{matrix}] \begin{matrix} \to & \cdot 4 & \to \\ \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 \\ \to & - I & \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} \to & + I I & \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}] \\ 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}

$\begin{align*}\frac{5x-2}{(2x-1)(3x+4)} &= \frac{A}{2x-1}+\frac{B}{3x+4}\\ 5x-2 &= A(3x+4)+B(2x-1)\\ 5x-2 &= 3Ax+4A+2Bx-B\\ 5 &= 3A+2B\\ -2 &= 4A-B\\ \begin{bmatrix} \begin{matrix} 3 & 2\\ 4 & -1 \end{matrix} \left| \begin{matrix} 5\\ -2 \end{matrix}\right. \end{bmatrix} \begin{matrix} \rightarrow & \cdot 4 & \rightarrow\\ \rightarrow & \cdot 3 & \rightarrow \end{matrix} & \begin{bmatrix} \begin{matrix} 12 & 8\\ 12 & -3 \end{matrix} \left| \begin{matrix} 20\\ -6 \end{matrix}\right. \end{bmatrix} \begin{matrix} & \rightarrow &\\ \rightarrow & -I & \rightarrow \end{matrix} \begin{bmatrix} \begin{matrix} 12 & 8\\ 0 & -11 \end{matrix} \left| \begin{matrix} 20\\ -26 \end{matrix}\right. \end{bmatrix} \begin{matrix} \rightarrow & \cdot 11 & \rightarrow\\ \rightarrow & \cdot 8 & \rightarrow \end{matrix} \begin{bmatrix} \begin{matrix} 132 & 88\\ 0 & -88 \end{matrix} \left| \begin{matrix} 220\\ -208 \end{matrix}\right. \end{bmatrix} \\ & \begin{matrix} \rightarrow & +II & \rightarrow\\ & \rightarrow & \end{matrix} \begin{bmatrix} \begin{matrix} 132 & 0\\ 0 & -88 \end{matrix} \left| \begin{matrix} 12\\ -208 \end{matrix}\right. \end{bmatrix} \begin{matrix} \rightarrow & \div 132 & \rightarrow\\ \rightarrow & \div -88 & \rightarrow \end{matrix} \begin{bmatrix} \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix} \left| \begin{matrix} \frac{1}{11}\\ \frac{26}{11} \end{matrix}\right. \end{bmatrix}\\ A = \frac{1}{11}, \ & B =\frac{26}{11}\\ \frac{5x-2}{(2x-1)(3x+4)} &= \frac{\frac{1}{11}}{2x-1}+\frac{\frac{26}{11}}{3x+4} \end{align*}$
::5x-2-2x-1(3x+4)=A2x-1+B3x+45x-2=A(3x+4)+B(2x-1)+B(2x-1)5x-2=3A+4A+2Bx-B5=3A-2B-2B-2=4A-B[324-1-1-5-5-5-2,5-2,44332}[1281212-3-3}[128-11-20-26][11-8}[132880-8_B3x+45x+45x-28=A(3x+4)]=A-2x-2B[324-4-1-1][3x+4]=11x-126x-1(3x+4)]=11x-116x-113x+4]

To confirm, add the fractions.

\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}

$\begin{align*}\frac{5x-2}{(2x-1)(3x+4)} &= \frac{\frac{1}{11}}{2x-1}+\frac{\frac{26}{11}}{3x+4}\\ 5x-2 &= \frac{1}{11}(3x+4)+\frac{26}{11}(2x-1)\\ 55x-22 &= 3x+4+26(2x-1)\\ 55x-22 &= 3x+4+52x-26\\ 55x-22 &= 55x-22\end{align*}$
::要确认,请添加分数。 5x-2-2-2x-1(3x+4)=1112x-1+26113x+45x-2=111(3x+4)+2611(2x-1)-55x-25x=3x-22=3x+4-26(2x-1)55x-22=3x-22x=3x+3x4+52x-26555x-22

Summary

Partial fraction decomposition is a procedure that reverses adding fractions with unlike denominators.
::部分碎片分解是一种程序,可以将与分母不同的分母反向增加分数。

Each individual factor of the denominator must be represented in the decomposition.
::分母的每一个个别因素都必须在分解中代表。

Linear factors raised to a power greater than one must have each successive power included as a separate denominator.
::向一个以上强国提出的线性因素必须包括作为单独分母的每一个相继权力。

Review
::回顾

Decompose the following rational expressions. Practice using matrices with at least one of the problems.
::将以下合理表达方式分解。使用至少有一个问题的矩阵练习。

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

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

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

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

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

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

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

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

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

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

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

12. Confirm your answer to #1 by adding the partial fractions.
::12. 通过添加部分分数确认您对 #1 的回答。

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

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。

章节大纲

常规

部分分数

Partial Fraction Decomposition ::分数分数分数分数分数分数分数

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)