配方x²-xy+y²-2x+5/2=(1/2)(2x²-2xy+2y²-4x+5)=(1/2)[(x²-2xy+y²)+(x²-4x+4)+y²+1]=(1/2)[(x-y)²+(x-2)²+y²+1]∵(x-y)²≥0,(x-2)²≥0,y²≥0...