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人気のある代数 >

3=a^9+2

解

3 = a^{9} + 2

解

a = 1, a = - \frac{1}{2} + \frac{3}{2} i, a = - \frac{1}{2} - \frac{3}{2} i, a = cos (\frac{2 π}{9}) + i sin (\frac{2 π}{9}), a = cos (\frac{8 π}{9}) + i sin (\frac{8 π}{9}), a = cos (\frac{14 π}{9}) + i sin (\frac{14 π}{9}), a = cos (\frac{2 π}{9}) - i sin (\frac{2 π}{9}), a = cos (\frac{4 π}{9}) + i sin (\frac{4 π}{9}), a = cos (\frac{10 π}{9}) + i sin (\frac{10 π}{9})

解答ステップ

3 = a^{9} + 2

辺を交換する

a^{9} + 2 = 3

2

を右側に移動します

a^{9} = 1

equationを

u = a^{3}

と以下で書き換える:

u^{3} = a^{9}

u^{3} = 1

解く

u^{3} = 1 : u = 1, u = \frac{- 1 + 3 i}{2}, u = \frac{- 1 - 3 i}{2}

u = 1, u = \frac{- 1 + 3 i}{2}, u = \frac{- 1 - 3 i}{2}

再び

u = a^{3}

に置き換えて以下を解く:

a

解く

a^{3} = 1 : a = 1, a = - \frac{1}{2} + \frac{3}{2} i, a = - \frac{1}{2} - \frac{3}{2} i

解く

a^{3} = \frac{- 1 + 3 i}{2} : a = cos (\frac{2 π}{9}) + i sin (\frac{2 π}{9}), a = cos (\frac{8 π}{9}) + i sin (\frac{8 π}{9}), a = cos (\frac{14 π}{9}) + i sin (\frac{14 π}{9})

解く

a^{3} = \frac{- 1 - 3 i}{2} : a = cos (\frac{2 π}{9}) - i sin (\frac{2 π}{9}), a = cos (\frac{4 π}{9}) + i sin (\frac{4 π}{9}), a = cos (\frac{10 π}{9}) + i sin (\frac{10 π}{9})

解答は

a = 1, a = - \frac{1}{2} + \frac{3}{2} i, a = - \frac{1}{2} - \frac{3}{2} i, a = cos (\frac{2 π}{9}) + i sin (\frac{2 π}{9}), a = cos (\frac{8 π}{9}) + i sin (\frac{8 π}{9}), a = cos (\frac{14 π}{9}) + i sin (\frac{14 π}{9}), a = cos (\frac{2 π}{9}) - i sin (\frac{2 π}{9}), a = cos (\frac{4 π}{9}) + i sin (\frac{4 π}{9}), a = cos (\frac{10 π}{9}) + i sin (\frac{10 π}{9})

グラフ

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