(x-y)²+(x-4)²+(y+2)²
＝(x-y)²/1+(4-x)²/1+(y+2)²/1
≥[(x-y)+(4-x)+(y+2)]²/(1+1+1)
＝12.
故所求最小值为：12.
此时,x-y＝4-x＝y+2,
即x＝2,y＝0.