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

Toán Lớp 6: Tìm x a/ x( $x^{2}$ + 4) = 0 b/ (3x – 2)( 4 + 7x) = 0 c/ $x^{2}$ – 8x = 0

Toán Lớp 6: Tìm x
a/ x( $x^{2}$ + 4) = 0
b/ (3x – 2)( 4 + 7x) = 0
c/ $x^{2}$ – 8x = 0

Comments ( 2 )

  1. Giải đáp:

     a, x(x^2+4)=0

    ⇒\(\left[ \begin{array}{l}x=0\\x^2+4=0\end{array} \right.\) 

    ⇒\(\left[ \begin{array}{l}x=0\\x^2=0-4\end{array} \right.\) 

    ⇒\(\left[ \begin{array}{l}x=0\\x^2=-4 (loại)\end{array} \right.\) 

    Vậy x= 0

    b, (3x – 2)( 4 + 7x) = 0

    ⇒\(\left[ \begin{array}{l}3x – 2=0\\4 + 7x = 0\end{array} \right.\)

    ⇒\(\left[ \begin{array}{l}3x=2\\7x=-4\end{array} \right.\)

    ⇒\(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{-4}{7}\end{array} \right.\)

    Vậy x∈{2/3 ; -4/7}

    c, x^2-8x=0

    ⇒x(x-8)=0

    ⇒\(\left[ \begin{array}{l}x=0\\x-8=0\end{array} \right.\) 

    ⇒\(\left[ \begin{array}{l}x=0\\x=0+8\end{array} \right.\) 

    ⇒\(\left[ \begin{array}{l}x=0\\x=8\end{array} \right.\) 

    Vậy x∈{0;8}

    #Hien

    Lời giải và giải thích chi tiết:

     

  2. a, x(x^2 + 4) = 0

    ->  $\left[\begin{matrix} x=0\\ x^2 + 4=0\end{matrix}\right.$

    -> $\left[\begin{matrix} x=0\\ x^2 =-4\text{  loại}\end{matrix}\right.$

    -> x = 0

    Vậy x = 0

    b, (3x – 2)(4 + 7x) = 0

    -> $\left[\begin{matrix} 3x-2=0\\ 4+7x=0\end{matrix}\right.$

    -> $\left[\begin{matrix} x=\frac{2}{3}\\ x=\frac{-4}{7}\end{matrix}\right.$

    Vậy $\left[\begin{matrix} x=\frac{2}{3}\\ x=\frac{-4}{7}\end{matrix}\right.$

    c, x^2 – 8x = 0

    -> x(x – 8) = 0

    -> $\left[\begin{matrix} x=0\\ x-8=0\end{matrix}\right.$

    -> $\left[\begin{matrix} x=0\\ x=8\end{matrix}\right.$

    Vậy $\left[\begin{matrix} x=0\\ x=8\end{matrix}\right.$

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

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