已知x(x+1)-(x²+y）=-3,求（x²＋y²）÷2-xy的值

已知x(x+1)-(x²+y）=-3,求（x²＋y²）÷2-xy的值
已知x(x+1)-(x²+y）=-3,求（x²＋y²）÷2-xy的值

已知x(x+1)-(x²+y）=-3,求（x²＋y²）÷2-xy的值
x(x+1)-(x²+y）=-3
x²+x-x²-y=-3
x-y=-3
所以原式=(x²+y²-2xy)/2
=(x-y)²/2
=9/2

x(x+1)-(x²+y）=-3
x²+x-x²-y=-3
∴x-y=-3

（x²＋y²）÷2-xy
=(x²-2xy+y²)÷2
=(x-y)²÷2
=(-3)²÷2
=4.5

x(x+1)-(x²+y）=-3,
x平方+x-x平方-y=-3
x-y=-3
（x²＋y²）÷2-xy
=[(x-y)^2+2xy]÷2-xy
=（9+2xy)÷2-xy
=9÷2
=4.5

x(x+1)-(x²+y)=x²+x-x²-y=x-y=-3
（x²＋y²）÷2-xy
=(x²+y²-2xy)÷2
=(x-y)²÷2
=9/2

展开得 x-y=-3, 求得4.5