(x+√3i)^3=log√2（下标）1/2^4 求x,√3i这是√乘于i。i^2=-1
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(x+√3i)^3=log√2（下标）1/2^4 求x,√3i这是√乘于i。i^2=-1
(x+√3i)^3=log√2（下标）1/2^4 求x,
√3i这是√乘于i。i^2=-1
(x+√3i)^3=log√2（下标）1/2^4 求x,√3i这是√乘于i。i^2=-1
log√2（下标）1/2^4
=log√2（下标）(1/√2)^8
=log√2（下标）(√2)^(-8)
=-8
(x+√3i)^3=-8
x+√3i=-2
x=-2-√3i
(x+√3i)^3=log√2 1/2^4
(x+√3i)^3=log2^(1/2) 2^(-4)
(x+√3i)^3=(-4)/(1/2)=-8=(-2)^3
x+√3i=-2
x=-2-√3i
log√2（下标）1/2^4
=lg1/2^4 /lg√2
=-4/(1/2)
=-8
(x+√3i)^3=-8=8(cosπ+isinπ)
所以x+√3i=2{cos[(2kπ+π)/3]+isin[(2kπ+π)/3]}
k=0,1,2
所以x+√3i=1+i√3
x+√3i=-2
x+√3i=1-i√3
x=1
x=-2-√3i
x=1-2√3i
解方程 (x+√3i)³=log‹√2›(1/2)⁴
(x+√3i)³=log₂(1/2) ⁸=log₂(2⁻⁸)= - 8
故得x+(√3)i=-2[cos(2kπ/3)+isin(2kπ/3)]，(k=0,1,2)
k=0时，x+(√3)i=-2,即x&#...
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解方程 (x+√3i)³=log‹√2›(1/2)⁴
(x+√3i)³=log₂(1/2) ⁸=log₂(2⁻⁸)= - 8
故得x+(√3)i=-2[cos(2kπ/3)+isin(2kπ/3)]，(k=0,1,2)
k=0时，x+(√3)i=-2,即x₁=-2-(√3)i
k=1时，x+(√3)i=-2[cos(2π/3)+isin(2π/3)]=-2[-cos(π/3)+isin(π/3)]=-2[-1/2+(√3/2)i]=1-(√3)i
故x₂=1-2(√3)i
k=2时，x+(√3)i=-2[cos(4π/3)+isin(4π/3)]=-2[-cos(π/3)-isin(π/3)]=-2[-1/2-(√3/2)i]=1+(√3)i
故x₃=1.
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