已知实数xy,满足10x²-2xy+y²+6x+1=0,求x+y
人气:118 ℃ 时间:2019-12-19 12:50:17
解答
10x²-2xy+y²+6x+1=0
(3x+1)²+(x-y)²=0
3x+1=0
x-y=0
所以
x=y=-1/3
x+y=-2/33x+1=x-y =3x+1=0x-y=0啊(3x+1)²+(x-y)²=0平方是≥0的,而右边=0所以只能是3x+1=0x-y=0(3x+1)²等于什么分解出来的?(x-y)等于。。。。。。。。10x²-2xy+y²+6x+1=0x²-2xy+y²+9x²+6x+1=0即可10x²拆成x²+9x²
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