直线与方程 (13 18:7:26)直线l1:x+y+a=0,l2:x+ay+1=0和l3:ax+y+1=0能构成三角形,求a的取值范围
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![直线与方程 (13 18:7:26)直线l1:x+y+a=0,l2:x+ay+1=0和l3:ax+y+1=0能构成三角形,求a的取值范围](/uploads/image/z/405791-71-1.jpg?t=%E7%9B%B4%E7%BA%BF%E4%B8%8E%E6%96%B9%E7%A8%8B+%2813+18%3A7%3A26%29%E7%9B%B4%E7%BA%BFl1%EF%BC%9Ax%2By%2Ba%3D0%2Cl2%EF%BC%9Ax%2Bay%2B1%3D0%E5%92%8Cl3%EF%BC%9Aax%2By%2B1%3D0%E8%83%BD%E6%9E%84%E6%88%90%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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