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

Toán Lớp 10: x^2-2x+3m-2=0 tìm m thỏa mãn x1<1

Toán Lớp 10: x^2-2x+3m-2=0 tìm m thỏa mãn x1<1

Comments ( 2 )

  1. $x_1<1<x_2$
    $⇒ \left\{ \begin{array}{l}x_1-1<0\\x_2-1>0\end{array} \right.$
    $⇒ (x_1-1)(x_2-1)<0$
    $⇔ x_1x_2-(x_1+x_2)+1<0$
    Theo định lý Vi-ét, ta có:
    $\left\{ \begin{array}{l}x_1+x_2=2\\x_1x_2=3m-2\end{array} \right.$
    Vậy $x_1x_2-(x_1+x_2)+1<0$
    $⇔ 3m-2-2+1<0$
    $⇔ 3m-3<0$
    $⇔ 3m<3$
    $⇔ m<1$
    Vậy $m<1$.
     

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