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

Toán Lớp 8: 1.Phân tích đa thức thành nhân tử a,x^4+4 b,x^4+x^2+1 2. Cho x>hoặc =2.C/m x^3+4x^2-3x-18>hoặc bằng 0

Toán Lớp 8: 1.Phân tích đa thức thành nhân tử
a,x^4+4
b,x^4+x^2+1
2. Cho x>hoặc =2.C/m x^3+4x^2-3x-18>hoặc bằng 0

Comments ( 2 )

  1. Giải đáp:
    $ 1. $
    a) x^4 + 4
    = x^4 + 4 + 4x^2 – 4x^2
    = (x^4 + 4x^2 + 4) – 4x^2
    = (x^2 + 2)^2 – (2x)^2
    = (x^2 + 2 – 2x)(x^2 + 2 + 2x)
    b) x^4 + x^2 + 1
    = x^4 – x^3 + x^3 + x^2 – x^2 + x^2 + x – x + 1
    = (x^4 – x^3 + x^2) + (x^3 – x^2 + x) + (x^2 – x + 1)
    = x^2 (x^2 – x + 1) + x(x^2 – x + 1) + (x^2 – x + 1)
    = (x^2 – x + 1)(x^2 + x + 1)
    $ 2. $
    x^3 + 4x^2 – 3x – 18
    = x^3 – 2x^2 + 6x^2 – 12x + 9x – 18
    = x^2 (x – 2) + 6x(x – 2) + 9(x- 2)
    = (x – 2)(x^2 + 6x + 9)
    = (x – 2)(x + 3)^2
    Vì (x + 3)^2 ≥ 0 $∀x$
    Mà x ≥ 2 → x – 2 ≥ 0
    ⇒ ( x – 2 ) ( x + 3 ) 2 ≥ 0    ( đ.p.c.m )

  2. $ 1) $
    a) x^4 + 4
    = x^4 + 4 + 4x^2 – 4x^2
    = (x^4 + 4x^2 + 4) – 4x^2
    = (x^2 + 2)^2 – (2x)^2
    = (x^2 + 2 – 2x)(x^2 + 2 + 2x)
    b) x^4 + x^2 + 1
    = x^4 – x^3 + x^3 + x^2 – x^2 + x^2 + x – x + 1
    = (x^4 – x^3 + x^2) + (x^3 – x^2 + x) + (x^2 – x + 1)
    = x^2 (x^2 – x + 1) + x(x^2 – x + 1) + (x^2 – x + 1)
    = (x^2 – x + 1)(x^2 + x + 1)
    $ 2) $
    Ta có:
    x^3 + 4x^2 – 3x – 18
    = x^3 – 2x^2 + 6x^2 – 12x + 9x – 18
    = x^2 (x – 2) + 6x(x – 2) + 9(x- 2)
    = (x – 2)(x^2 + 6x + 9)
    = (x – 2)(x + 3)^2
    Vì (x + 3)^2 ≥ 0 $∀x$
    Mà x ≥ 2 → x – 2 ≥ 0
    => (x – 2)(x + 3)^2 ≥ 0 $(đpcm)$

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