Toán Lớp 8: chứng minh x^8 + x^4 + 1 chia hết cho x^2 + x + 1 = 2 cách

Question

Toán Lớp 8:  chứng minh x^8 + x^4 + 1 chia hết cho x^2 + x + 1 = 2 cách

cobelolen · Answer

Giải đáp +Lời giải và giải thích chi tiết:     C1:    x^8 + x^4 + 1   = x^8 + x^4 + 1 + x^2 - x^2  + x -  x     = (x^8 - x^2) + (x^4 - x) + (x^2 + x + 1)     = x^2(x^6 - 1) + x(x^3 - 1) + (x^2 + x +1)   = x^2[(x^3)^2 - 1^2] + x(x^3 - 1) + (x^2 + x + 1)   =x^2(x^3-1)(x^3+1) + x(x^3 + 1) + (x^2 + x + 1)     = x^2(x^3 + 1)(x - 1)(x^2 + x + 1) + x(x^2-1)(x^2 + x + 1)  + (x^2 + x + 1)    M&agrave;  x^2 + x + 1    vdots  x^2 + x + 1   -> x^2(x^3 + 1)(x - 1)(x^2 + x + 1) + x(x^2-1)(x^2 + x + 1)  + (x^2 + x + 1)    vdots  x^2 + x + 1   (đpcm)    C2:     x^8 + x^4 + 1   = x^8 + 2x^4 - x^4 + 1    = (x^8 + 2x^4   +1 ) - x^4     = [(x^4)^2 - 2. x^4 . 1 + 1^2] - (x^2)^2     = (x^4 + 1)^2 - (x^2)^2   = [(x^4  + 1 ) - x^2][(x^4 + 1) + x^2]    = (x^4 + 1  -x^2)(x^4 + 1 + x^2)     = (x^4 - x^2 + 1)(x^4 + 2x^2 - x^2 + 1)   = (x^4 - x^2 + 1)[(x^4 + 2x^2 + 1)- x^2]    = (x^4 - x^2 + 1)[(x^2 + 1)^2  - x^2]     = (x^4 - x^2 + 1)(x^2 + 1 - x)(x^2 + x + 1 )   vdots   x^2 + x + 1

tonhitruc · Answer

~ gửi bạn ~   x^8 + x^4 + 1   = x^8 + 2x^4  + 1- x^4   = (x^4 + 1)^2 - x^4   = (x^4 + 1 + x^2)(x^4 + 1 - x^2)   = (x^4 + 2^2x + 1 - x^2)(x^4 - x^2 + 1)   = [(x^2 + 1)^2 - x^2](x^4 - x^2 + 1)   = (x^2 + x - x)(x^2 + x + x)(x^4 - x^2 + 1)  ⋮ x^2 + x + 1   ____________________________________   x^8 + x^4 + 1   = x^8 - x^2 + x^4 - x + x^2 + x + 1   = x^2(x^6 - 1) + x(x^3 - 1) + (x^2 + x + 1)   = x^2(x^3 + 1)(x - 1)(x^2 + x + 1)  + x(x^2 - 1)(x^2 + x + 1) + (x^2 + x + 1) ⋮ x^2 + x + 1