2x^2+1= 1/((x^2)-4)

解

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

x = \frac{7 + 89}{2}, x = - \frac{7 + 89}{2}, x = \frac{7 - 89}{4}, x = - \frac{7 - 89}{4}

解答ステップ

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

簡素化

\frac{1}{( x ^{2} ) - 4} : \frac{1}{x ^{2} - 4}

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

以下で両辺を乗じる:

x^{2} - 4

2 x^{2} (x^{2} - 4) + x^{2} - 4 = 1

解く

2 x^{2} (x^{2} - 4) + x^{2} - 4 = 1 : x = \frac{7 + 89}{2}, x = - \frac{7 + 89}{2}, x = \frac{7 - 89}{4}, x = - \frac{7 - 89}{4}

x = \frac{7 + 89}{2}, x = - \frac{7 + 89}{2}, x = \frac{7 - 89}{4}, x = - \frac{7 - 89}{4}

解を検算する

未定義の (特異) 点を求める:

x = 2, x = - 2

未定義のポイントを解に組み合わせる:

x = \frac{7 + 89}{2}, x = - \frac{7 + 89}{2}, x = \frac{7 - 89}{4}, x = - \frac{7 - 89}{4}