Given that the image of the function y = | x2-1 | X-1 has exactly two intersections with the image of the function y = KX, then the value range of the real number k is Function y= |x2-1| x-1 = |x+1| • |x-1| x-1 = x+1 ,x＞1 -(x+1) ,-1≤ x＜1 x+1 ,x＜-1 x+1 ,x＞1 -(x+1) ,-1≤ x＜1 x+1 ,x＜-1 How did you get here?

Given that the image of the function y = | x2-1 | X-1 has exactly two intersections with the image of the function y = KX, then the value range of the real number k is Function y= |x2-1| x-1 = |x+1| • |x-1| x-1 = x+1 ,x＞1 -(x+1) ,-1≤ x＜1 x+1 ,x＜-1 x+1 ,x＞1 -(x+1) ,-1≤ x＜1 x+1 ,x＜-1 How did you get here?

Is X-1 in the denominator
The key is to determine the sign in the absolute value by the range of X,
Then the absolute value sign is removed, and finally the denominator is reduced
When x > 1, x + 1 > 0, X-1 > 0, so there is
(|x+1||x-1|)/(x-1)=(x+1)(x-1)/(x-1)=x+1
When - 1 ≤ x ≤ 1, x + 1 ≥ 0, X-1 ≤ 0, so there is
(|x+1||x-1|)/(x-1)=(x+1)[-(x-1)]/(x-1)=-(x+1)
When x