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Binomial form of the 84-year Mathematics College Entrance Examination
m = x |+ 1/| x |-2)3 =[(| x |+ 1/| x |)-2]3。 Let the constant term be r term, then,

c3(r)*[(|x|+ 1/|x|)|x|,^(3-r)]*(-2)^r

Let the constant term in [(| x |+1| x|) (3-r)] be the k term, then

c(3-r)(k)* {(|x|)^(3-r-k)}*(|x|)^(-k)=c(3-r)(k)* {(|x|)^(3-r-2*k)}

Let 3-r-2*K=0, r can be 0, 1, 2,3, then k corresponds to 1.5, 1, 0.5,0.

Bring r= 1, K= 1 into the formula M=- 12, and add r=3, K=0, M=-8, and then get the original formula =- 12-8=-20.

I think I made it clearer than the one below.