First of all, answer your question about assignment: the purpose of assignment is to find the sum of coefficients. If x=0, a0 is on the right, but the sum of other coefficients cannot be obtained. If x= 1, and A0+A 1+...+A5 is on the right, the required coefficient sum can be obtained. Are you clear? Not "I can assign 0, I can assign 1."

Let f (x) = (1-2x) 5 = A0+a1x+a2x2+...+a5x5.

Then f (1) = A0+a1+...+A5 = (1-2) 5 =-1,A0 = C (5 5,0)15 =/kloc-0.

So a 1+a2...+a5=-2.

F (- 1) = A0-A 1+...+A5 = ( 1+2) 5 = 243。

So f (1)+f (-1) = 2 (A0+A2+A4) = 242, and f (1)-f (-1) = 2 (A1+A3.

Therefore, (a1+a3+a5) (A0+a2+a4) =-122 *1=-14762.

|a 1|+|a2|+...| a5 | =-(a 1+a3+a5)+(a2+a4+A0)-A0 = 122+ 12 1- 1 = 242

Whether it is X (x-2/x) 7 or [x (x-2/x)] 7 is different for people who study mathematics. Make it clear, ask questions to let others know the exact meaning of the question clearly, and don't let people guess. Of course, this question does not affect the result.

The sum is f (1) =1(1-2) 7 =-1.