Let a and B satisfy a 2-8a + 6 = 0 and 6B 2-8b + 1 = 0, and find the value of AB + 1ab

Let a and B satisfy a 2-8a + 6 = 0 and 6B 2-8b + 1 = 0, and find the value of AB + 1ab

As 6b2-8b2-8b + 1 = 0, then B ≠ 0, then (1b) 2 − 8 × 1b + 6 = 0, when a ≠ 1b, then a, 1b is the two roots of the equation x2-8x + 6 = 0, let's set X1 = a, X2 = 1b, then X1 + x2 = 8, X1 = 6, so (1b) 2 − 2 (1b) 2 {(1b) 2 − 8 × 1b + 6 + 6 = 0, and (1b) 2 {(1b) 2 {(1b) 2 {(1b) 2 {(1b) 2 {(1b) 2, that is ab = 1, so AB + 1ab = 2.in summary: when a ≠ 1b, AB + 1A = 263, AB + 1b = 263; when a = 1b = 1b = 1b, AB + 1b = 263; when a = 1b = 1b = 1b = 1b = 1b, AB + 1A = 263; ab + 1A = 1b = 1b = 1b = AB + 1