初中数学题
已知 x(x-1)-(x^2-y)=-3, 求(x^2+y^2)÷2-xy 的值
x{x-1}-{x^2-y}=-3 即x^2-x-x^2+y=-3 -x+y=-3 x-y=3 而x^2+y^2=(x-y)^2+2xy 所以{x^2+y^2}/2 -xy =[(x-y)^2+2xy]/2-xy =(x-y)^2/2+xy-xy =(x-y)^2/2 =9/2
解:由题意: x(x-1)-(x^2-y)=-x+y=-3 ∴x-y=3 ∴x^2+y^2=(x-y) ^2+2xy=3^2+2xy=9+2xy ∴原式= (9+2xy)/2-xy =9/2+xy-xy =9/2
我做出来了~~~~ 由x(x-1)-(x^2-y)=-3得 y-x=-3平方得 y^2+x^2-2xy=9 xy=x^2+y^2-9/2 代入(x^2+y^2)÷2-xy 得 (x^2+y^2)÷2-xy =9/2
解:由题意: x(x-1)-(x^2-y)=-x+y=-3 即:x-y=3 则: (x^2+y^2)÷2-xy =(x-y)^2/2=9/2
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