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

Toán Lớp 8: Tìm x a) -x^3+9x^2-27x+27=0 b) (x+2)^3+(x-3)^3=0 c) (x-1)^2+(x-2)^2=0

Toán Lớp 8: Tìm x
a) -x^3+9x^2-27x+27=0
b) (x+2)^3+(x-3)^3=0
c) (x-1)^2+(x-2)^2=0

Comments ( 2 )

  1. Giải đáp:
    a. -x^3+9x^2-27x+27=0
    <=> -(x^3-9x^2+27x-27)=0
    <=> x^2-9x^2+27x-27=0
    <=> (x-3)^3=0
    <=> x=3
    $\\$
    b. (x+2)^3+(x-3)^3=0
    <=> x^3+6x^2+12x+8+x^3-9x^2+27x-27=0
    <=> 2x^3-3x^2+39x-19=0
    <=> 2x^3-x^2-2x^2+x+38x-19=0
    <=> (2x-1)(x^2-x+19)=0
    <=>$\left[ \begin{array}{l}2x-1=0\\x^2-x+19=0\end{array} \right.$
    <=>$\left[ \begin{array}{l}x=\dfrac{1}{2}\\x\notin\mathbb{R}\end{array} \right.$
    <=> x=1/2
    $\\$
    c. (x-1)^2+(x-2)^2=0
    <=> x^2-2x+1+x^2-4x+4=0
    <=> 2x^2-6x+5=0
    <=> x\notin \mathbb{R}

  2. Giải đáp:
     
    Lời giải và giải thích chi tiết:
     a) -x³+9x²-27x+27=0
    ⇒-(x³ – 9x²+27x-27)=0
    ⇒ -(x³ – 3.x².3+3.x.3²-3³)=0
    ⇒ -(x-3)³=0
    ⇒ -x+3=0
    ⇒-x=-3
    ⇒x=3
    Vậy x=3
    b)(x+2)³+(x-3)³=0
    ⇒ (x+2+x-3)[(x+2)²-(x+2)(x+3)+(x-3)²]=0
    ⇒ (2x-1)[(x²+4x+4)-(x²+3x+2x+6)+(x²-6x+9)]=0
    ⇒(2x-1)(x²+4x+4-x²-3x-2x-6+x²-6x+9)=0
    ⇒(2x-1)(x²-x+19)=0
    ⇒\(\left[ \begin{array}{l}2x-1=0\\x²-x+19=0\end{array} \right.\) 
    ⇒\(\left[ \begin{array}{l}x=1/2\\x∉ R\end{array} \right.\) 
    Vậy x=1/2
    c) (x-1)²+(x-2)²=0
    ⇒x²-2x+1+x²-4x+4=0
    ⇒2x²-6x+5=0
    ⇒x∉R
    ⇒x=∅
    Vậy x = ∅

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