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xiiiiiiiiii

y√[(x-y)y]=-√[(x-3)y] 故y≤0,x-3≤0 原式 =√(x^2-8x+16)+√(y-1)^2-√(x-3)^2 =|x-4|+|y-1|-|x-3| =4-x+1-y-(3-x) =2-y

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