The absolute value of a times X-1 is greater than 2 + a

Is this the case? Is this what is the case? The square can get a & \\124;x-1-x-1 (x-1) \\|x-1 |x-1-1 \124;124;x-1? Is this the case? The square can get a \\\\\\\124;\|\|\| 178; (x-1; (178; (x-1) \\\\\\\\\\\\\\\\\\124; 178; (178; (178; (178= - B

The inequality with absolute value holds fast!
|X-1 | + | X-5 | > A is constant, find the value range and solution process of a!

Method 1: expansion method when x ≤ 1, | X-1 | + | X-5 | = 1-x + 5-x = 6-2x, the minimum value is obtained at x = 1, the minimum value is 4, when 1 ＜ x ＜ 5, | X-1 | + | X-5 | = x - 1 + 5-x = 4, the minimum value is 4, when x ≥ 5, | X-1 | + | X-5 | = X-1 + X-5 = 2x - 6, the minimum value is obtained at x = 5, the minimum value is 4

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The absolute value of a times X-1 is greater than 2 + a