1 ---- x(x-1)-(x^2-y)=-2,求(x^2+y^2)/2-xy的值
问题描述：
1 ---- x(x-1)-(x^2-y)=-2,求(x^2+y^2)/2-xy的值
2 ---- 试说明不论x,y取什么有理数,多项式x^2+y^2-2x+2y+3的值总是正数.
^2大家都知道的吧,就是平方,
waiting.
答
1.
∵x(x-1)-(x^2-y)=-2
∴x^2-x-x^2+y=-2
x-y=2
∵(x-y)^2=x^2+y^2-2xy=4
∴x^2+y^2=4+2xy
∴(x^2+y^2)/2-xy=(4+2xy)/2-xy=2+xy-xy=2
2.x^2+y^2-2x+2y+3=x^2-2x+1+y^2+2y++1+1
=(x-1)^2+(y+1)^2+1
∵(x-1)^2、(y+1)^2均大于等于0
∴x^2+y^2-2x+2y+3=(x-1)^2+(y+1)^2+1>0
请选我吧````(^-^)