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

Toán Lớp 8: a) x3 – 9x =0 b) x^2 -4 + (x-2)^2 =0 c) -x^2 + 5x -6 = 0

Toán Lớp 8: a) x3 – 9x =0
b) x^2 -4 + (x-2)^2 =0
c) -x^2 + 5x -6 = 0

Comments ( 2 )

  1. a)
    x^3 – 9x  = 0
    => x (x^2-9)=0
    =>x (x^2-3^2)=0
    =>x(x-3)(x+3)=0
    =>x=0 hoặc x-3=0 hoặc x+3=0
    +)x=0
    +)x-3=0=>x=3
    +)x+3=0=>x=-3
    Vậy x \in {0;3;-3}
    b)
    x^2 – 4 + (x-2)^2 = 0
    => (x^2-2^2) + (x-2)^2=0
    =>(x-2)(x+2) + (x-2)^2=0
    =>(x-2)(x+2+x-2)=0
    =>(x-2) . 2x=0
    =>x-2=0 hoặc x=0
    +)x=0
    +)x-2=0=>x=2
    Vậy x \in {0;2}
    c)
    -x^2 + 5x – 6 = 0
    => -x^2 + 3x + 2x – 6 =0
    => -x (x-3) + 2 (x-3) =0
    =>(2-x)(x-3)=0
    =>2-x=0 hoặc x-3=0
    +)2-x=0=>x=2
    +)x-3=0=>x=3
    Vậy x \in{2;3}

  2. a)x^3-9x=0
    <=>x(x^2-9)=0
    <=>x(x-3)(x+3)=0
    <=>\(\left[ \begin{array}{l}x=0\\x-3=0\\x+3=0\end{array} \right.\) 
    <=>\(\left[ \begin{array}{l}x=0\\x=3\\x=-3\end{array} \right.\) 
    Vậy x=0 hoặc x=3 hoặc x=-3
    b)x^2-4+(x+2)^2=0
    <=>(x-2)(x+2)+(x+2)(x+2)=0
    <=>(x+2)(x-2+x+2)=0
    <=>2x(x+2)=0
    <=>\(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\) 
    Vậy x=0 hoặc x=-2
    c)-x^2+5x-6=0
    <=>-(x^2-5x+6)=0
    <=>x^2-3x-2x+6=0
    <=>x(x-3)-2(x-3)=0
    <=>(x-2)(x-3)=0
    <=>\(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) 
    Vậy  x=2 hoặc x=3

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

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