It is proved that (ABC + BCD + CDA + DAB) ^ 2 - (AB CD) (BC DA) (CA BD) = ABCD (a + B + C + D) ^ 2

It is proved that (ABC + BCD + CDA + DAB) ^ 2 - (AB CD) (BC DA) (CA BD) = ABCD (a + B + C + D) ^ 2

In fact, this is a simple calculation. Expand item by item. It's OK to merge. As long as you have the patience to do it step by step, but you need to know how to expand each item. It's not difficult to mention it. It's inconvenient to answer it step by step

What is y = ax + B / CX + D? How can we get the center (- D / C, a / C)?

Y = (AX + b) / (Cx + D) denotes hyperbolic equation in analytic geometry. When C ≠ 0 is required to find the center, the original equation can be transformed identically. Y = (AX + b) / (Cx + D) = [(A / C) (x + B / a)] / (x + D / C) = (A / C) [1 - (AD BC) / (AC)] / (x + D / C) = A / C - [(AD BC) / C & # 178;] / (x + D / C) makes (ad BC) / C & # 178

Known ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = (X-2) ^ 4, evaluation: 1, a + B + C + D + E 2, B + D process write clearly

Expand (X-2) ^ 4, that is, (X-2) ^ 4 = x ^ 4-8x ^ 3 + 24x ^ 2-32x + 16, so a = 1b = - 8C = 24D = - 32e = 16 A + B + C + D + e = 1b + D = - 40