
已知x-2y=3,x²-2xy+4y²=11,求下列各值; (1)xy (2)x²-2xy²
急,在线等答案。顺便帮我把因式分解也解决一下:-x²+4x-44x³-4²y-(x-y)谢谢!...
急,在线等答案。顺便帮我把因式分解也解决一下:-x²+4x-4 4x³-4²y-(x-y)
谢谢! 展开
谢谢! 展开
1个回答
展开全部
x-2y=3,
两边平方,
x²-4xy+4y²=9 (1)
x²-2xy+4y²=11 (2)
两式相减,
2xy=2
xy=1
x²y-2xy² 这题是不是少了y
=xy(x-2y)
=1×3
=3
因式分解:
-x²+4x-4
=-(x²-4x+4)
=-(x-2)²
4x³-4x²y-(x-y)
=4x²(x-y)-(x-y)
=(x-y)(4x²-1)
=(x-y)(2x+1)(2x-1)
两边平方,
x²-4xy+4y²=9 (1)
x²-2xy+4y²=11 (2)
两式相减,
2xy=2
xy=1
x²y-2xy² 这题是不是少了y
=xy(x-2y)
=1×3
=3
因式分解:
-x²+4x-4
=-(x²-4x+4)
=-(x-2)²
4x³-4x²y-(x-y)
=4x²(x-y)-(x-y)
=(x-y)(4x²-1)
=(x-y)(2x+1)(2x-1)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询