## 1. 問題の内容

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1. 問題の内容

この問題は、与えられた $x$ と $y$ の値、または $x$ の値を用いて、様々な式の値を求める問題です。具体的には、次の問題が含まれています。

1. $x = \frac{\sqrt{5}+2}{\sqrt{5}-2}$、$y = \frac{\sqrt{5}-2}{\sqrt{5}+2}$ のとき、$x+y$、$xy$、$x^2y+xy^2$、$x^2+y^2$、$x^3+y^3$ を求めよ。

2. $x = \sqrt{2}-1$ のとき、$x+\frac{1}{x}$、$x^2+\frac{1}{x^2}$、$x^3+\frac{1}{x^3}$を求めよ。

3. $x = 1-\sqrt{5}$ のとき、$x^2-2x-4$ を求めよ。

4. $\frac{\sqrt{2}}{\sqrt{2}-1}$ の整数部分を $a$、小数部分を $b$ とするとき、$a$、$b$、$a+b+b^2$ を求めよ。

5. $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ を計算せよ。

6. $x < 3$ のとき、$\sqrt{x^2-6x+9}$ を $x$ の多項式で表せ。

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2. 解き方の手順

### 問題1

まず、$x$ と $y$ を簡単にします。

$x = \frac{\sqrt{5}+2}{\sqrt{5}-2} = \frac{(\sqrt{5}+2)^2}{(\sqrt{5}-2)(\sqrt{5}+2)} = \frac{5 + 4\sqrt{5} + 4}{5 - 4} = 9 + 4\sqrt{5}$

$y = \frac{\sqrt{5}-2}{\sqrt{5}+2} = \frac{(\sqrt{5}-2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)} = \frac{5 - 4\sqrt{5} + 4}{5 - 4} = 9 - 4\sqrt{5}$

(1) $x+y = (9 + 4\sqrt{5}) + (9 - 4\sqrt{5}) = 18$

(2) $xy = (9 + 4\sqrt{5})(9 - 4\sqrt{5}) = 81 - 16 \cdot 5 = 81 - 80 = 1$

(3) $x^2y + xy^2 = xy(x+y) = 1 \cdot 18 = 18$

(4) $x^2 + y^2 = (x+y)^2 - 2xy = 18^2 - 2 \cdot 1 = 324 - 2 = 322$

(5) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = 18(18^2 - 3 \cdot 1) = 18(324 - 3) = 18 \cdot 321 = 5778$

### 問題2

(1) $\frac{1}{x} = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$

$x+\frac{1}{x} = (\sqrt{2}-1) + (\sqrt{2}+1) = 2\sqrt{2}$

(2) $x^2 + \frac{1}{x^2} = (x+\frac{1}{x})^2 - 2 = (2\sqrt{2})^2 - 2 = 8 - 2 = 6$

(3) $x^3 + \frac{1}{x^3} = (x+\frac{1}{x})^3 - 3(x+\frac{1}{x}) = (2\sqrt{2})^3 - 3(2\sqrt{2}) = 16\sqrt{2} - 6\sqrt{2} = 10\sqrt{2}$

### 問題3

$x = 1 - \sqrt{5}$ なので、

$x^2 - 2x - 4 = (1-\sqrt{5})^2 - 2(1-\sqrt{5}) - 4 = (1 - 2\sqrt{5} + 5) - 2 + 2\sqrt{5} - 4 = 6 - 2\sqrt{5} - 6 + 2\sqrt{5} = 0$

### 問題4

$\frac{\sqrt{2}}{\sqrt{2}-1} = \frac{\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{2 + \sqrt{2}}{2 - 1} = 2 + \sqrt{2}$

$\sqrt{2} \approx 1.414$ より、$2 + \sqrt{2} \approx 3.414$

(1) したがって、$a = 3$

(2) $b = (2 + \sqrt{2}) - 3 = \sqrt{2} - 1$

(3) $a + b + b^2 = 3 + (\sqrt{2}-1) + (\sqrt{2}-1)^2 = 3 + \sqrt{2} - 1 + (2 - 2\sqrt{2} + 1) = 3 + \sqrt{2} - 1 + 3 - 2\sqrt{2} = 5 - \sqrt{2}$

### 問題5

$\frac{1}{1+\sqrt{2}-\sqrt{3}} = \frac{1}{(1+\sqrt{2})-\sqrt{3}} \cdot \frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2})+\sqrt{3}} = \frac{1+\sqrt{2}+\sqrt{3}}{(1+\sqrt{2})^2 - 3} = \frac{1+\sqrt{2}+\sqrt{3}}{1 + 2\sqrt{2} + 2 - 3} = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2} - 0} = \frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}} = \frac{(1+\sqrt{2}+\sqrt{3})\sqrt{2}}{2\sqrt{2}\sqrt{2}} = \frac{\sqrt{2}+2+\sqrt{6}}{4}$

### 問題6

$x < 3$ より、$x - 3 < 0$

$\sqrt{x^2-6x+9} = \sqrt{(x-3)^2} = |x-3| = -(x-3) = 3-x$

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3. 最終的な答え

1. (1) 18 (2) 1 (3) 18 (4) 322 (5) 5778

2. (1) $2\sqrt{2}$ (2) 6 (3) $10\sqrt{2}$

3. 0

4. (1) 3 (2) $\sqrt{2} - 1$ (3) $5 - \sqrt{2}$

5. $\frac{\sqrt{2}+2+\sqrt{6}}{4}$

6. $3-x$