解由P(x,y)=x^2+2xy+2y^2-y+1
=x^2+2xy+y^2+y²-y+1
=（x+y）²+（y-1/2)²+3/4≥3/4
即P(x,y)≥3/4
当当y=-1，x=-1时P(x,y)有最大值=（x+y）²+（y-1/2)²+3/4=（-2）²+(-1-1/2）²+3/4=7
即二元多项式P(x,y)=x^2+2xy+2y^2-y+1的取值范围是[3/4,7]
P(x,y)=x^2+2xy+2y^2-y+1
3/4<=（x+y）^2+(y-1/2)^2+3/4<=7 (4+9/4+3/4)
即 3/4<=P(x,y)<=7
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