（1）联立两个方程通分,得到：x^2 - 6x + k = 0
所以x1 + x2 = 6,又2x1 - x2 = 6,所以可以解得：
x1 = 4,x2 = 2,带回方程得到k = -4+12 = 8；
（2）A(4,2),B(2,4),所以可以发现OAB为一个等腰三角形,(3,3)为AB中点,面积即为(3√2×2√2)/2 = 6
Copyright © 1996 - 2007 Paldohunch,All Rights Reserved
（1）联立两个方程通分,得到：x^2 - 6x + k = 0
所以x1 + x2 = 6,又2x1 - x2 = 6,所以可以解得：
x1 = 4,x2 = 2,带回方程得到k = -4+12 = 8；
（2）A(4,2),B(2,4),所以可以发现OAB为一个等腰三角形,(3,3)为AB中点,面积即为(3√2×2√2)/2 = 6
Copyright © 1996 - 2007 Paldohunch,All Rights Reserved