解答
25x(10x12)=(25x10)x12
解答
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)
求解步骤
25x(10x12)=(25x10)x12
将 (25x10)x12para o lado esquerdo
25x⋅10x12−25x10x12=0
因式分解 25x⋅10x12−25x10x12:−25x13(x−910)(x2+910x+1092)(x6+310x3+1032)
−25x13(x−910)(x2+910x+1092)(x6+310x3+1032)=0
使用零因数法则: If ab=0then a=0or b=0x=0orx−910=0orx2+910x+1092=0orx6+310x3+1032=0
解 x−910=0:x=910
解 x2+910x+1092=0:x=−92565+i231092,x=−92565−i231092
解 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)
解为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)