文[1]的例6及其“正解”如下:题目函数y=(m-1)xm-1+(m-3)x+1,当m为何值时,它是一次函数.解当m-1=0且m-3≠0时,为一次函数.解得m=1;当m-1=0且m-3≠0时,为一次函数.解得m=±1;当m-1=1且(m-1)+(m-3)≠0时,为一次函数.解得m=-2.所以当m=±1或m=-2时,它是一次函数.评论这个“正解”不对!当m=1时,y=(1-1)x1-1+(1-3)x+1,即y=0x0-2x+1,即y=-2x+1(x≠0).它不是一次函数!它的图像不是一条直 Example 6 of [1] and its “positive solution” are as follows: Problem Function y = (m -1) xm -1 + (m -3) x +1, which is a linear function when m is the value. When m-1 = 0 and m-3 ≠ 0, it is a linear function, and m = 1; when m-1 = 0 and m-3 ≠ 0, it is a linear function. When m-1 = 1 and (m-1) + (m-3) ≠ 0, it is a linear function, solving m = -2, so it is a linear function when m = ± 1 or m = -2. When m = 1, y = (1-1) x1-1 + (1-3) x + 1, that is, y = 0x0-2x + 1, that is, y = -2x + 1 (x ≠ 0). It is not a linear function! Its image is not straight