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

Toán Lớp 8: $x^2( x – 6 ) + 18 – 3x = 0$ Có bao nhiều giá trị x thõa mãn

Toán Lớp 8: $x^2( x – 6 ) + 18 – 3x = 0$
Có bao nhiều giá trị x thõa mãn

Comments ( 2 )

  1. \(x^2(x-6)+18-3x=0\\\Leftrightarrow x^2(x-6)-3(-6+x)=0\\\Leftrightarrow (x-6)(x^2-3)=0\\\Leftrightarrow(x-6)(x-\sqrt3)(x+\sqrt3)=0\\\Leftrightarrow \left[ \begin{array}{l}x-6=0\\x-\sqrt3=0\\x+\sqrt3=0\end{array} \right.\\\Leftrightarrow\left[ \begin{array}{l}x=6\\x=\sqrt3\\x=-\sqrt3\end{array} \right.\)
     

  2. ~ Bạn tham khảo ~
    x^2(x-6) + 18-3x=0
    <=> x^2(x-6) + 3(6-x) = 0
    <=> x^2(x-6) – 3(x-6) = 0
    <=> (x-6)(x^2-3)=0
    <=>[(x-6=0),(x^2-3=0):}
    <=>[(x=6),(x=+-\sqrt(3)):}
    Vậy x={6;+-\sqrt(3)}

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

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