已知x(x+1)-(x²+y)=-3,求(x²+y²)÷2-xy的值
人气:459 ℃ 时间:2019-08-22 00:17:06
解答
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
推荐
- 已知x+y=3,xy=2,求 x²+xy+y²的值和(x-y)²
- (1)已知x+y=6,x²-y²=5,求x-y的值(2)若x+y=3,x²+y²=5,求xy的值
- 已知x+y=5,xy=3 (1)x²+y²的值 (2)求x-y(x>y)的值
- 已知x+y=3,xy=1,求:(1)x²+y²(2)(x-y)²的值
- 已知(a+b)²=7,(a-b)²=3,求(1)x²+y² (2)xy的值
- 已知x1,x2是一元二次方程3x*x+2x-6=0的两个根,不解方程,求x1*x1+x1x2+x2*x2和x2/x1+x1/x2的值
- 为什么是how much does it weigh 而不是how many..
- Changes took place in the Scottish universities which had a major impact on the Society.For example
猜你喜欢