Let f (x) satisfy AF (x) + BF (1-x) = C / x, a, B and C are constants, and the absolute values a and B are unequal, then f (x) is obtained

Let f (x) satisfy AF (x) + BF (1-x) = C / x, a, B and C are constants, and the absolute values a and B are unequal, then f (x) is obtained

Let f (x) satisfy AF (x) + BF (1-x) = C / x, where a, B and C are constants and absolute value a ≠ absolute value B, then f (x)

Because
AF (x) + BF (1-x) = C / X formula one
that
AF (1-x) + BF (x) = C / (1-x) formula 2
Formula a - formula B
（a²-b²）f（x）=c【a/(x）-b/(1-x)】
f(x)=c【a/(x）-b/(1-x)】/(a²-b²）

If the equation f ^ 2 (x) + BF (x) + C = 0 of X has five real roots, where f (x) = x + 1 / | x |, write the value range of B and the functional relationship between B and C

When x > 0, f (x) = x + 1 / X decreases monotonically in the interval (0,1), increases monotonically in the interval (1, + ∞), f (x) min = f (1) = 2. When x 2, there are three intersections between the line y = f (x) and the image f (x), so when f (x) = 2, there are two intersections between the line y = f (x) and the image f (x), so when f (x) n), then M > 2, n = 2 △ = B & # 178; - 4 * 1 * C = B & # 178; -