由已知得:△ECF周长=x+y+ √(x²+y²) =2,
所以(√(x²+y²)=2-(x+y)
x²+y²=[2-(x+y)]²
x²+y²=4-4(x+y)+(x+y)²
x²+y²=4-4(x+y)+x²+2xy+y²
4-4(x+y)+2xy=0
2-2(x+y)+xy=0
2(x+y)-xy=2
即:-xy=2-(x+y)
打的没有分行,所以有点看不来.
由已知得:△ECF周长=x+y+ √(x²+y²) =2,
所以(√(x²+y²)=2-(x+y)
x²+y²=[2-(x+y)]²
x²+y²=4-4(x+y)+(x+y)²
x²+y²=4-4(x+y)+x²+2xy+y²
4-4(x+y)+2xy=0
2-2(x+y)+xy=0
2(x+y)-xy=2
即:-xy=2-(x+y)
打的没有分行,所以有点看不来.