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

Toán Lớp 8: `m`, `x + 6x^2 =0` `n` , `(x +1)=(x+1)^2 ` Tìm x

Tháng Chín 8, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: m, x + 6x^2 =0
n , (x +1)=(x+1)^2
Tìm x

Comments ( 2 )

  1. tuyetnhungthu
    tuyetnhungthu
    8 Tháng Chín, 2022 at 9:46 chiều
    Reply
    m) $x+6x^2=0$
    $⇔ x(1+6x)=0$
    $⇔ \left[ \begin{array}{l}x=0\\1+6x=0\end{array} \right.$
    $⇔ \left[ \begin{array}{l}x=0\\x=-\dfrac{1}{6}\end{array} \right.$
    n) $(x+1)=(x+1)^2$
    $⇔ (x+1)^2-(x+1)=0$
    $⇔ (x+1)(x+1-1)=0$
    $⇔ (x+1).x=0$
    $⇔ \left[ \begin{array}{l}x+1=0\\x=0\end{array} \right.$
    $⇔ \left[ \begin{array}{l}x=-1\\x=0\end{array} \right.$
     

  2. mocmien87
    mocmien87
    8 Tháng Chín, 2022 at 9:46 chiều
    Reply
    Giải đáp:
    m) $x+6x^2=0\\\Rightarrow x.(1+6x)=0\\\Rightarrow\left[\begin{matrix}x=0\\1+6x=0\end{matrix} \right.\\\Rightarrow\left[\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix} \right.$ 
    Vậy $x\in\left\{0;\dfrac{-1}{6}\right\}$.
    n) $x+1=(x+1)^2\\\Rightarrow (x+1)^2-(x+1)=0\\\Rightarrow (x+1).\!(x+1-1)=0\\\Rightarrow \left[\begin{matrix}x+1=0\\x=0\end{matrix}\right.\\\Rightarrow \left[\begin{matrix}x=-1\\x=0\end{matrix}\right.$
    Vậy $x\in\left\{-1;0\right\}$.

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