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

Toán Lớp 8: PHân tích đa thức thành nhân tử: x^3-2x=3x^2-6 Cíuuuuu mik zới

Toán Lớp 8: PHân tích đa thức thành nhân tử:
x^3-2x=3x^2-6
Cíuuuuu mik zới

Comments ( 2 )

  1. @Chery
    x^3 – 2x = 3x^2 – 6
    => x^3 – 2x – 3x^2 + 6 = 0
    => ( x^3 – 2x ) – ( 3x^2 – 6 ) = 0
    => x ( x^2 – 2 ) – 3 ( x^2 – 2 ) = 0
    => ( x^2 – 2 )( x – 3 ) = 0
    => $\left \{ {{x^2 – 2 = 0} \atop {x – 3 = 0}} \right.$ 
    => $\left \{ {{x^2 = 2 } \atop {x – 3 = 0}} \right.$ 
    => $\left \{ {{x = \sqrt{2} ; – \sqrt{2} } \atop {x = 3}} \right.$ 
    Vậy x $\in$ { $\sqrt{2}$ ; – $\sqrt{2}$ ; 3 }
     

  2. x^3 – 2x = 3x^2 – 6
    <=> x^3 – 2x – 3x^2 + 6 = 0
    <=> x^3 – 3x^2 – 2x + 6 =0
    <=> x^3 – 2x – 3x^2 + 6 =0
    <=> x(x^2 – 2) – 3(x^2-2) = 0
    <=> (x-3)(x^2 – 2) = 0
    <=> \(\left[ \begin{array}{l}x-3=0\\x^2-2=0\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=3\\x^2 = 2\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=3\\x=±\sqrt{2}\end{array} \right.\) 
    Vậy x∈{3;±\sqrt{2}}

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