:∵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)-(5...