作业帮正方形OABC.ADEF的顶点A.D.C在坐标轴上,点F在AB上,点B.E在函数Y=1/x(x>0)的图象上,则点E的坐标是A:[(√5+1)/2,(√5-1)/2]B:[(3+√5)/2,(3-√5)/2]C:[(√5-1)/2,(√5+1)/2]D:[(3-√5)/2,(3+√5)/2]
来源：学生作业帮助网 编辑：作业帮 时间：2024/08/09 19:02:17
![正方形OABC.ADEF的顶点A.D.C在坐标轴上,点F在AB上,点B.E在函数Y=1/x(x>0)的图象上,则点E的坐标是A:[(√5+1)/2,(√5-1)/2]B:[(3+√5)/2,(3-√5)/2]C:[(√5-1)/2,(√5+1)/2]D:[(3-√5)/2,(3+√5)/2]](/uploads/image/z/12083929-25-9.jpg?t=%E6%AD%A3%E6%96%B9%E5%BD%A2OABC.ADEF%E7%9A%84%E9%A1%B6%E7%82%B9A.D.C%E5%9C%A8%E5%9D%90%E6%A0%87%E8%BD%B4%E4%B8%8A%2C%E7%82%B9F%E5%9C%A8AB%E4%B8%8A%2C%E7%82%B9B.E%E5%9C%A8%E5%87%BD%E6%95%B0Y%3D1%2Fx%28x%3E0%29%E7%9A%84%E5%9B%BE%E8%B1%A1%E4%B8%8A%2C%E5%88%99%E7%82%B9E%E7%9A%84%E5%9D%90%E6%A0%87%E6%98%AFA%3A%5B%28%E2%88%9A5%2B1%29%2F2%2C%28%E2%88%9A5-1%29%2F2%5DB%3A%5B%283%2B%E2%88%9A5%29%2F2%2C%283-%E2%88%9A5%29%2F2%5DC%3A%5B%28%E2%88%9A5-1%29%2F2%2C%28%E2%88%9A5%2B1%29%2F2%5DD%3A%5B%283-%E2%88%9A5%29%2F2%2C%283%2B%E2%88%9A5%29%2F2%5D)
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正方形OABC.ADEF的顶点A.D.C在坐标轴上,点F在AB上,点B.E在函数Y=1/x(x>0)的图象上,则点E的坐标是A:[(√5+1)/2,(√5-1)/2]B:[(3+√5)/2,(3-√5)/2]C:[(√5-1)/2,(√5+1)/2]D:[(3-√5)/2,(3+√5)/2]
正方形OABC.ADEF的顶点A.D.C在坐标轴上,点F在AB上,点B.E在函数Y=1/x(x>0)的图象上,则点E的坐标是
A:[(√5+1)/2,(√5-1)/2]
B:[(3+√5)/2,(3-√5)/2]
C:[(√5-1)/2,(√5+1)/2]
D:[(3-√5)/2,(3+√5)/2]
正方形OABC.ADEF的顶点A.D.C在坐标轴上,点F在AB上,点B.E在函数Y=1/x(x>0)的图象上,则点E的坐标是A:[(√5+1)/2,(√5-1)/2]B:[(3+√5)/2,(3-√5)/2]C:[(√5-1)/2,(√5+1)/2]D:[(3-√5)/2,(3+√5)/2]
应该是点B和E在函数y=1/x的图像上,没有S的.
设正方形OABC边长为a,正方形ADEF边长为b.
则B（a,a）,E（a+b,b）
代入一次函数,得出a=1,b=(√5-1)/2
所以B(1,1),E( √5/2+1/2,√5/2-1/2).