解
25x(10x12)=(25x10)x12
解
x=0,x=910,x=−92565+i231092,x=−92565−i231092,x=64232(3⋅532+232(45)32)cos(92π)+i64232(3⋅532+232(45)32)sin(92π),x=64232(3⋅532+232(45)32)cos(98π)+i64232(3⋅532+232(45)32)sin(98π),x=64232(3⋅532+232(45)32)cos(914π)+i64232(3⋅532+232(45)32)sin(914π),x=64232(3⋅532+232(45)32)cos(92π)−i64232(3⋅532+232(45)32)sin(92π),x=64232(3⋅532+232(45)32)cos(94π)+i64232(3⋅532+232(45)32)sin(94π),x=64232(3⋅532+232(45)32)cos(910π)+i64232(3⋅532+232(45)32)sin(910π)
解答ステップ
25x(10x12)=(25x10)x12
(25x10)x12を左側に移動します
25x⋅10x12−25x10x12=0
因数 25x⋅10x12−25x10x12:−25x13(x−910)(x2+910x+1092)(x6+310x3+1032)
−25x13(x−910)(x2+910x+1092)(x6+310x3+1032)=0
零因子の原則を使用:ab=0ならば a=0または b=0x=0orx−910=0orx2+910x+1092=0orx6+310x3+1032=0
解く x−910=0:x=910
解く x2+910x+1092=0:x=−92565+i231092,x=−92565−i231092
解く x6+310x3+1032=0:x=64232(3⋅532+232(45)32)cos(92π)+i64232(3⋅532+232(45)32)sin(92π),x=64232(3⋅532+232(45)32)cos(98π)+i64232(3⋅532+232(45)32)sin(98π),x=64232(3⋅532+232(45)32)cos(914π)+i64232(3⋅532+232(45)32)sin(914π),x=64232(3⋅532+232(45)32)cos(92π)−i64232(3⋅532+232(45)32)sin(92π),x=64232(3⋅532+232(45)32)cos(94π)+i64232(3⋅532+232(45)32)sin(94π),x=64232(3⋅532+232(45)32)cos(910π)+i64232(3⋅532+232(45)32)sin(910π)
解答はx=0,x=910,x=−92565+i231092,x=−92565−i231092,x=64232(3⋅532+232(45)32)cos(92π)+i64232(3⋅532+232(45)32)sin(92π),x=64232(3⋅532+232(45)32)cos(98π)+i64232(3⋅532+232(45)32)sin(98π),x=64232(3⋅532+232(45)32)cos(914π)+i64232(3⋅532+232(45)32)sin(914π),x=64232(3⋅532+232(45)32)cos(92π)−i64232(3⋅532+232(45)32)sin(92π),x=64232(3⋅532+232(45)32)cos(94π)+i64232(3⋅532+232(45)32)sin(94π),x=64232(3⋅532+232(45)32)cos(910π)+i64232(3⋅532+232(45)32)sin(910π)