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

Toán Lớp 7: (x-3)(x+2) <0 và (x-1)(x+4/3) >0

Toán Lớp 7: (x-3)(x+2) <0 và (x-1)(x+4/3) >0

Comments ( 2 )

  1. a,
    (x-3)(x+2)<0
    $⇔\left[\begin{matrix} \begin{cases} x-3>0\\x+2<0 \end{cases}\\ \begin{cases} x-3<0\\x+2>0\\ \end{cases} \end{matrix}\right.$
    $⇔\left[\begin{matrix} \begin{cases} x>3\\x< -2 \end{cases}\\ \begin{cases} x<3\\x > -2\\ \end{cases} \end{matrix}\right.$
    $⇔\begin{cases} x<3\\x > -2 \end{cases}$
    Vậy: x∈{-1;0;1;2}
    b, 
    (x-1)(x+\frac{4}{3})>0
    $⇔\left[\begin{matrix} \begin{cases} x-1>0\\x+\dfrac{4}{3}>0 \end{cases}\\ \begin{cases} x-1<0\\x+\dfrac{4}{3}<0\\ \end{cases} \end{matrix}\right.$
    $⇔\left[\begin{matrix} \begin{cases} x>1\\x> -\dfrac{4}{3} \end{cases}\\ \begin{cases} x<1 \\x< -\dfrac{4}{3}\\ \end{cases}  \end{matrix}\right.$
    $⇔\begin{cases} x>1\\x < -\dfrac{4}{3} \end{cases}$
    Vậy: x>1;x < -\frac{4}{3}

  2. (x-3)(x+2) <0
    ⇒\(\left[ \begin{array}{l}\left \{ {{x-3>0} \atop {x+2<0}} \right.\\\left \{ {{x-3<0} \atop {x+2>0}} \right.\end{array} \right.\) 
    ⇒\(\left[ \begin{array}{l}\left \{ {{x>3} \atop {x<-2}} \right.\\\left \{ {{x<3} \atop {x>-2}} \right.\end{array} \right.\) 
    ⇒\(\left[ \begin{array}{l}x ko có giá trị \\-2<x<3\end{array} \right.\)
    ⇒x∈{-1;0;1;2}
    (x-1)(x+4/3) >0
    ⇒\(\left[ \begin{array}{l}\left \{ {{x-1>0} \atop {x+\frac{4}{3}>0}} \right.\\\left \{ {{x-1<0} \atop {x+\frac{4}{3}<0}} \right.\end{array} \right.\) 
     ⇒\(\left[ \begin{array}{l}\left \{ {{x>1} \atop {x>\frac{-4}{3}}} \right.\\\left \{ {{x<1} \atop {x<\frac{-4}{3}}} \right.\end{array} \right.\) 
    ⇒\(\left[ \begin{array}{l}x>1\\x=\frac{-4}{3}\end{array} \right.\)

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

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