$2x + \frac{1}{2x} = \sqrt{7}$のとき、次の式の値を求める問題です。 (1) $4x^2 + \frac{1}{4x^2}$ (2) $8x^3 + \frac{1}{8x^3}$ (3) $64x^6 + \frac{1}{64x^6}$

1. 問題の内容

$2x + \frac{1}{2x} = \sqrt{7}$のとき、次の式の値を求める問題です。

(1) $4x^2 + \frac{1}{4x^2}$

(2) $8x^3 + \frac{1}{8x^3}$

(3) $64x^6 + \frac{1}{64x^6}$

2. 解き方の手順

(1)

$2x + \frac{1}{2x} = \sqrt{7}$の両辺を2乗します。

$(2x + \frac{1}{2x})^2 = (\sqrt{7})^2$

$4x^2 + 2(2x)(\frac{1}{2x}) + \frac{1}{4x^2} = 7$

$4x^2 + 2 + \frac{1}{4x^2} = 7$

$4x^2 + \frac{1}{4x^2} = 7 - 2$

$4x^2 + \frac{1}{4x^2} = 5$

(2)

$2x + \frac{1}{2x} = \sqrt{7}$の両辺を3乗します。

$(2x + \frac{1}{2x})^3 = (\sqrt{7})^3$

$(2x)^3 + 3(2x)^2(\frac{1}{2x}) + 3(2x)(\frac{1}{2x})^2 + (\frac{1}{2x})^3 = 7\sqrt{7}$

$8x^3 + 3(4x^2)(\frac{1}{2x}) + 3(2x)(\frac{1}{4x^2}) + \frac{1}{8x^3} = 7\sqrt{7}$

$8x^3 + 6x + \frac{3}{2x} + \frac{1}{8x^3} = 7\sqrt{7}$

$8x^3 + \frac{1}{8x^3} + 3(2x + \frac{1}{2x}) = 7\sqrt{7}$

$8x^3 + \frac{1}{8x^3} + 3\sqrt{7} = 7\sqrt{7}$

$8x^3 + \frac{1}{8x^3} = 7\sqrt{7} - 3\sqrt{7}$

$8x^3 + \frac{1}{8x^3} = 4\sqrt{7}$

(3)

(1)で求めた、$4x^2 + \frac{1}{4x^2} = 5$の両辺を3乗します。

$(4x^2 + \frac{1}{4x^2})^3 = 5^3$

$(4x^2)^3 + 3(4x^2)^2(\frac{1}{4x^2}) + 3(4x^2)(\frac{1}{4x^2})^2 + (\frac{1}{4x^2})^3 = 125$

$64x^6 + 3(16x^4)(\frac{1}{4x^2}) + 3(4x^2)(\frac{1}{16x^4}) + \frac{1}{64x^6} = 125$

$64x^6 + 12x^2 + \frac{3}{4x^2} + \frac{1}{64x^6} = 125$

$64x^6 + \frac{1}{64x^6} + 3(4x^2 + \frac{1}{4x^2}) = 125$

$64x^6 + \frac{1}{64x^6} + 3 \times \frac{3}{4} = 125$

$64x^6 + \frac{1}{64x^6} + \frac{9}{4} = 125$

$64x^6 + \frac{1}{64x^6} = 125-12=113$

$64x^6 + \frac{1}{64x^6} = 125 - \frac{3}{4}\times 3 = 113.25$

$64x^6 + \frac{1}{64x^6} = 125 - 3\times \frac{5}{4} = 125- \frac{15}{4}$

$64x^6 + \frac{1}{64x^6} = \frac{500-15}{4}$

$64x^6 + \frac{1}{64x^6} = \frac{485}{4}$

3. 最終的な答え

(1) $4x^2 + \frac{1}{4x^2} = 5$

(2) $8x^3 + \frac{1}{8x^3} = 4\sqrt{7}$

(3) $64x^6 + \frac{1}{64x^6} = \frac{485}{4}$