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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: `(x – 1)(x^3+ x^2+ x + 1)` Thực hiện phép tính

Toán Lớp 8: (x – 1)(x^3+ x^2+ x + 1)
Thực hiện phép tính

Comments ( 2 )

  1. $(x-1)(x^3+x^2+x+1)\\=(x-1)[(x^3+x^2)+(x+1)]\\=(x-1)[x^2(x+1)+(x+1)]\\=(x-1)(x^2+1)(x+1)\\=[(x-1)(x+1)](x^2+1)\\=(x^2-1)(x^2+1)\\=x^4-1$
    Vậy $(x-1)(x^3+x^2+x+1)=x^4-1\,$
     

  2. Giải đáp:
    Cách 1:
    (x – 1)(x^3 + x^2 + x + 1)
    = x^4 + x^3 + x^2 + x – x^3 – x^2 – x – 1
    = x^4 + (x^3 – x^3) + (x^2 – x^2) + (x – x) – 1
    = x^4 – 1
    Cách 2: 
    (x – 1)(x^3 + x^2 + x + 1)
    = (x – 1)[(x^3 + x^2) + (x + 1)]
    = (x – 1)[x^2(x + 1) + (x + 1)]
    = (x – 1)(x + 1)(x^2 + 1)
    = (x^2 – 1)(x^2 + 1)
    = (x^2)^2 – 1^2
    = x^4 – 1
    Học tốt

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222-9+11+12:2*14+14 = ? ( )

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