數與座標系/4 有理數與無理數-範例

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範例11

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範例11類題1

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範例11類題2

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範例12

  1. 已知：\left\{ \begin{align} & x\in Q \\ & \left( x^{2}-1 \right)+\left( x^{2}-2x-3 \right)\times \sqrt{5}=0 \\ \end{align} \right.，求解：x\,
  2. 已知：\left\{ \begin{align} & \sqrt{11+\sqrt{72}}=a+b \\ & a\in N \\ & 0<b<1 \\ \end{align} \right.，求解：\frac{2a-1}{a+b+1}+\frac{7}{a-b-1}

答案

  1. x=-1\,
  2. \frac{2a-1}{a+b+1}+\frac{7}{a-b-1}=4

詳解

1

  1. x\in Q\Rightarrow \left\{ \begin{align} & \left( x^{2}-1 \right)\in Q \\ & \left( x^{2}-2x-3 \right)\in Q \\ \end{align} \right.
  2. \left\{ \begin{align} & \left( x^{2}-2x-3 \right)\in Q \\ & \left( x^{2}-2x-3 \right)\times \sqrt{5}\in Q \\ \end{align} \right.\Rightarrow \left( x^{2}-2x-3 \right)=0
  3. \left\{ \begin{matrix} \left( x^{2}-1 \right)\in Q \\ \left( x^{2}-1 \right)+0\times \sqrt{5}=0 \\ \end{matrix} \right.\Rightarrow \left( x^{2}-1 \right)=0
  4. \left\{ \begin{align} & \left( x^{2}-1 \right)=0 \\ & \left( x^{2}-2x-3 \right)=0 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & \left( x+1 \right)\left( x-1 \right)=0 \\ & \left( x-3 \right)\left( x+1 \right)=0 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & x=-1or0 \\ & x=-1or3 \\ \end{align} \right.\Rightarrow x=-1\#

2

  1. a+b=\sqrt{11+\sqrt{72}}=\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}
    =\sqrt{3^{2}+2\times 3\times \sqrt{2}+\sqrt{2}^{2}}=\sqrt{\left( 3+\sqrt{2} \right)^{2}}=3+\sqrt{2}
  2. \left\{ \begin{align} & a+b=3+\sqrt{2} \\ & a\in N \\ & 0<b<1 \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & b\notin Q \\ & 1<\sqrt{2}<2\Rightarrow 1-1<\sqrt{2}-1<2-1 \\ \end{align} \right.
    \Rightarrow 0<\sqrt{2}-1<1\Rightarrow \left\{ \begin{align} & b=\sqrt{2}-1 \\ & a=3+\sqrt{2}-b=4 \\ \end{align} \right.
  3. \frac{2a-1}{a+b+1}+\frac{7}{a-b-1}=\frac{8-1}{4+\left( \sqrt{2}-1 \right)+1}+\frac{7}{4-\left( \sqrt{2}-1 \right)-1}
    =\frac{7}{4+\sqrt{2}}+\frac{7}{4-\sqrt{2}}=\frac{7\left( 4-\sqrt{2} \right)+7\left( 4+\sqrt{2} \right)}{\left( 4+\sqrt{2} \right)\left( 4-\sqrt{2} \right)}=\frac{56}{16-2}=\frac{56}{14}=4\#

範例12類題1

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範例12類題2

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