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

Toán Lớp 8: c) (2x – 1)(x^2 – x + 1) = 2x^3 – 3x^2 + 2 d) (x + 1)(x + 2)(x + 5) – x^3 – 8x^2 + 7 = 0

Toán Lớp 8: c) (2x – 1)(x^2 – x + 1) = 2x^3 – 3x^2 + 2
d) (x + 1)(x + 2)(x + 5) – x^3 – 8x^2 + 7 = 0

Comments ( 2 )

  1. c) (2x – 1)(x^2 – x + 1) = 2x^3 – 3x^2 + 2
    <=>2x(x^2 – x + 1)-1(x^2 – x + 1)=2x^3 – 3x^2 + 2
    <=>2x^3-2x^2+2x-x^2+x-1=2x^3 – 3x^2 + 2
    <=>2x^2-2x^3-2x^2-x^2+3x^2+x+2x=1+2
    <=>3x=3
    <=>x=1
    Vậy S={1}
    d) (x + 1)(x + 2)(x + 5) – x^3 – 8x^2 + 7 = 0
    <=>[x(x+2)+1(x+2)](x + 5) – x^3 – 8x^2 + 7 = 0
    <=>(x^2+2x+x+2)(x + 5) – x^3 – 8x^2 + 7 = 0
    <=>(x^2+3x+2)(x+5) -x^3 – 8x^2 + 7 = 0
    <=>x^2(x+5)+3x(x+5)+2(x+5)-x^3 – 8x^2 + 7 = 0
    <=>x^3+5x^2+3x^2+15x+2x+10-x^3 – 8x^2 + 7 = 0
    <=>x^3+8x^2+17x+10-x^3 – 8x^2 + 7 = 0
    <=>17x=-10-7
    <=>17x=-17
    <=>x=-1
    Vậy S={-1}

  2. #Rùa
    Giải đáp+Lời giải và giải thích chi tiết:
     c)(2x-1)(x^2-x+1)=2x^3-3x^2+2
    ⇔ 2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0
    ⇔ 3x-3=0
    ⇔ 3x=3
    ⇔ x=1
    Vây S={1}
    d)(x+1)(x+2)(x+5)-x^3-8x^2+7=0
    ⇔ (x^2+2x+x+2)(x+5)-x^3-8x^2+7=0
    ⇔ (x^2+3x+2)(x+5)-x^3-8x^2+7=0
    ⇔ x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2+7=0
    ⇔ 17x+17=0
    ⇔ 17x=-17
    ⇔ x=-1
    Vậy S={-1}

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

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