多項式200811071636

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目錄

屬性

  • 資源類別:題目
  • 科目:數學
  • 主題:多項式
  • 次主題:多項式的四則運算, 除法原理
  • 摘要:f(x)=(x-b/a)*q(x)+r, 除法原理
  • 適用年級:10-12
  • 日期:2008/11/07
  • 編輯者:User:HsiaoCH
  • 參考資料:
  • 相關技術:
  • 題目狀態:

求解

  • 已知:\left\{ \begin{align}
  & f\left( x \right)\text{ is a polynomial} \\ 
 & \text{a,b}\in R \\ 
 & a\ne 0 \\ 
 & f\left( x \right)=\left( x-\frac{b}{a} \right)\times q\left( x \right)+r \\ 
\end{align} \right.
    1. f\left( x \right)=\left( ax-b \right)\times q_{1}\left( x \right)+r_{1} ,求 q_{1},r_{1}\,
    2. f\left( \frac{x}{a} \right)=\left( x-b \right)\times q_{2}\left( x \right)+r_{2} ,求 q_{2},r_{2}\,
    3. x\times f\left( x \right)=\left( ax-b \right)\times q_{3}\left( x \right)+r_{3} ,求 q_{3},r_{3}\,

答案

  1. \left\{ \begin{align}
  & q_{1}=\frac{q\left( x \right)}{a} \\ 
 & r_{1}=r \\ 
\end{align} \right.
  2. \left\{ \begin{align}
  & q_{2}=\frac{q\left( \frac{x}{a} \right)}{a} \\ 
 & r_{2}=r \\ 
\end{align} \right.
  3. \left\{ \begin{align}
  & q_{3}=\frac{q\left( x \right)\times x+r}{a} \\ 
 & r_{3}=\frac{b\times r}{a} \\ 
\end{align} \right.

詳解

1

  1. f\left( x \right)=\left( x-\frac{b}{a} \right)\times q\left( x \right)+r\Rightarrow a\times f\left( x \right)=a\times \left[ \left( x-\frac{b}{a} \right)\times q\left( x \right)+r \right]
    =a\times \left( x-\frac{b}{a} \right)\times q\left( x \right)+a\times r\Rightarrow a\times f\left( x \right)=\left( ax-b \right)\times q\left( x \right)+a\times r
    \Rightarrow f\left( x \right)=\frac{\left( ax-b \right)\times q\left( x \right)+a\times r}{a}=\left( ax-b \right)\times \frac{q\left( x \right)}{a}+r
  2. \left\{ \begin{align}
  & f\left( x \right)=\left( ax-b \right)\times \frac{q\left( x \right)}{a}+r \\ 
 & f\left( x \right)=\left( ax-b \right)\times q_{1}\left( x \right)+r_{1} \\ 
\end{align} \right.\Rightarrow \left\{ \begin{align}
  & q_{1}=\frac{q\left( x \right)}{a}\# \\ 
 & r_{1}=r\# \\ 
\end{align} \right.

2

  1. f\left( x \right)=\left( x-\frac{b}{a} \right)\times q\left( x \right)+r\Rightarrow f\left( \frac{x}{a} \right)=\left( \frac{x}{a}-\frac{b}{a} \right)\times q\left( \frac{x}{a} \right)+r
    =\left( \frac{x-b}{a} \right)\times q\left( \frac{x}{a} \right)+r=\left( x-b \right)\times \frac{q\left( \frac{x}{a} \right)}{a}+r
  2. \left\{ \begin{align}
  & f\left( \frac{x}{a} \right)=\left( x-b \right)\times \frac{q\left( \frac{x}{a} \right)}{a}+r \\ 
 & f\left( \frac{x}{a} \right)=\left( x-b \right)\times q_{2}\left( x \right)+r_{2} \\ 
\end{align} \right.\Rightarrow \left\{ \begin{align}
  & q_{2}=\frac{q\left( \frac{x}{a} \right)}{a}\# \\ 
 & r_{2}=r\# \\ 
\end{align} \right.

3

  1. f\left( x \right)=\left( ax-b \right)\times \frac{q\left( x \right)}{a}+r\Rightarrow x\times f\left( x \right)=\left( ax-b \right)\times \frac{q\left( x \right)\times x}{a}+r\times x
    =\left( ax-b \right)\times \frac{q\left( x \right)\times x}{a}+\left( ax-b \right)\times \frac{r}{a}+\frac{b\times r}{a}
    =\left( ax-b \right)\times \frac{q\left( x \right)\times x+r}{a}+\frac{b\times r}{a}
  2. \left\{ \begin{align}
  & x\times f\left( x \right)=\left( ax-b \right)\times \frac{q\left( x \right)\times x+r}{a}+\frac{b\times r}{a} \\ 
 & x\times f\left( x \right)=\left( ax-b \right)\times q_{3}\left( x \right)+r_{3} \\ 
\end{align} \right.\Rightarrow \left\{ \begin{align}
  & q_{3}=\frac{q\left( x \right)\times x+r}{a}\# \\ 
 & r_{3}=\frac{b\times r}{a}\# \\ 
\end{align} \right.
Facts about 多項式200811071636RDF feed
主題 多項式  +
摘要 f(x)=(x-b/a)*q(x)+r  +, and 除法原理  +
日期 2008年11月7日 (星期五)  +
次主題 多項式的四則運算  +, and 除法原理  +
科目 數學  +
編輯者 HsiaoCH  +
資源類別 題目  +
適用年級 10-12  +