2.设x,y为实数,且(x/1+i)*(y/1-2i)=5/1-3i,求x+y.

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2.设x,y为实数,且(x/1+i)*(y/1-2i)=5/1-3i,求x+y.
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2.设x,y为实数,且(x/1+i)*(y/1-2i)=5/1-3i,求x+y.
2.设x,y为实数,且(x/1+i)*(y/1-2i)=5/1-3i,求x+y.

2.设x,y为实数,且(x/1+i)*(y/1-2i)=5/1-3i,求x+y.
:∵a/(1-i)-b/(1-2i)=5/(1-3i)
左端通分得:
∴(a-2ai-b+bi)/(1-2i-i+2i^2)=5/(1-3i)
∴[(a-b)-(2a-b)i]/(-1-3i)=5/(1-3i)
-5-15i=(a-b)-3i(a-b)-3(2a-b)-(2a-b)i
根据实部等于实部,虚部等于虚部得:
-5-15i=(-5a+2b)-(5a-4b)i
∴2b-5a=-5.15=5a-4b
∴-2b=10,b=-5
a=(2b+5)/5=-5/2
x+y=-15/2