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

Toán Lớp 8: tìm x: $x^{3}$ – $2x^{2}$ – 7x = $x^{2}$ + 3x

Toán Lớp 8: tìm x:
$x^{3}$ – $2x^{2}$ – 7x = $x^{2}$ + 3x

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết !
    to Tìm x:
    x^3-2x^2-7x = x^2+3x
    <=> x^3-2x^2-7x-x^2-3x = 0
    <=> x^3+(-2x^2-x^2)+(-7x-3x) = 0
    <=> x^3-3x^2-10x = 0
    <=> x(x^2-3x-10) = 0
    <=> x(x^2-5x+2x-10) = 0
    <=> x[(x^2-5x)+(2x-10)] = 0
    <=> x[x(x-5)+2(x-5)] = 0
    <=> x(x+2)(x-5) = 0
    <=> \(\left[ \begin{array}{l}x=0\\x+2=0\\x-5=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=0\\x=-2\\x=5\end{array} \right.\) 
    Vậy S= {0; -2; 5}
     

  2. x^3 – 2x^2 – 7x = x^2 + 3x
    <=> x^3 – 2x^2 – 7x – x^2 – 3x = 0
    <=> x^3 – 3x^2 – 10x = 0
    <=> x^3 – 5x^2 + 2x^2 – 10x = 0
    <=> x^2(x – 5) + 2x(x – 5) = 0
    <=> (x^2 + 2x)(x – 5) = 0
    <=> x(x + 2)(x – 5) = 0
    <=> \(\left[ \begin{array}{l}x = 0 \\ x + 2 = 0 \\ x – 5 = 0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x = 0 \\ x =-2\\ x =5\end{array} \right.\) 
    Vậy x \in {0 ; -2 ; 5}
     

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

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