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

Toán Lớp 8: (x ² -9) ² = (x – 3 ) ² tìm x nha

Toán Lớp 8: (x ² -9) ² = (x – 3 ) ² tìm x nha

Comments ( 2 )

  1. (x^2-9)^2=(x-3)^2
    ⇔(x^2-9)^2-(x-3)^2=0
    ⇔(x^2-9-x+3)(x^2-9+x-3)=0
    ⇔(x^2-x-6)(x^2+x-12)=0
    ⇔[(x^2-3x)+(2x-6)][(x^2-3x)+(4x-12)]=0
    ⇔[x(x-3)+2(x-3)][x(x-3)+4(x-3)]=0
    ⇔(x-3)(x+2)(x-3)(x+4)=0
    ⇔(x-3)^2(x+2)(x+4)=0
    ⇔\(\left[ \begin{array}{l}x-3=0\\x+2=0\\x+4=0\end{array} \right.\) 
    ⇔\(\left[ \begin{array}{l}x=3\\x=-2\\x=-4\end{array} \right.\) 
    Vậy S={3,-2,-4}
     

  2. Giải đáp + Lời giải và giải thích chi tiết !
    to Tìm x:
    (x^2-9)^2 = (x-3)^2
    <=> (x^2-9)^2-(x-3)^2 = 0
    <=> (x^2-9-x+3)(x^2-9+x-3) = 0
    <=> (x^2-x-6)(x^2+x-12) = 0
    <=> (x^2-3x+2x-6)(x^2+4x-3x-12) = 0
    <=> [(x^2-3x)+(2x-6)][(x^2+4x)-(3x+12)] = 0
    <=> [x(x-3)+2(x-3)][x(x+4)-3(x+4)] = 0
    <=> (x+2)(x-3)(x-3)(x+4) = 0
    <=> (x+2)(x-3)^2(x+4) = 0
    <=> \(\left[ \begin{array}{l}x+2=0\\(x-3)^2=0\\x+4=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=-2\\x-3=0\\x=-4\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=-2\\x=3\\x=-4\end{array} \right.\) 
    Vậy S= {-2; 3; -4}
     

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