先把分式化简:
x/(1-i)上下同时乘以(1+i)得(x(1+i))/2
y/(1-2i)上下同时乘以(1+2i)得(y(1+2i))/5
5/(1-3i)上下同时乘以(1+3i)得(1+3i)/2
所以(x(1+i))/2+(y(1+2i))/5=(1+3i)/2
方程两边乘以10得:
5x(1+i)+2y(1+2i)=5(1+3i)
化简后得:
5x+2y-5+5xi+4yi-15i=0
所以得出方程组:
5x+2y-5=0
5x+4y-15=0
所以x=-1,y=5
先把分式化简:
x/(1-i)上下同时乘以(1+i)得(x(1+i))/2
y/(1-2i)上下同时乘以(1+2i)得(y(1+2i))/5
5/(1-3i)上下同时乘以(1+3i)得(1+3i)/2
所以(x(1+i))/2+(y(1+2i))/5=(1+3i)/2
方程两边乘以10得:
5x(1+i)+2y(1+2i)=5(1+3i)
化简后得:
5x+2y-5+5xi+4yi-15i=0
所以得出方程组:
5x+2y-5=0
5x+4y-15=0
所以x=-1,y=5