Tìm x,biết: a) (3-x)(x+2)>0 b) (x-2)(x+2020)>0

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1. Đáp án:

    

   Giải thích các bước giải:

   a) (3-x)(x+2)>0

   $\left \{ {{3-x>0} \atop {x+2>0}} \right.$

   $\left \{ {{x<3} \atop {x>-2}} \right.$

   ⇒-2<x<3

   b,(x-2)(x+2020)>0

   $\left \{ {{x-2>0} \atop {x+2020>0}} \right.$

   $\left \{ {{x>2} \atop {x>-2020}} \right.$

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2. ` a) ` ` (3 – x)(x + 2) > 0 `

   ` => ` \(\left[ \begin{array}{l}\left \{ {{3-x>0} \atop {x+2>0}} \right. \\\left \{ {{3-x<0} \atop {x+2<0}} \right. \end{array} \right.\) 

   ` => ` \(\left[ \begin{array}{l}\left \{ {{x>3} \atop {x>-2}} \right. \\\left \{ {{x<3} \atop {x<-2}} \right. \end{array} \right.\) 

   ` => ` \(\left[ \begin{array}{l}x>3\\x<-2\end{array} \right.\) 

   ` b) ` ` (x – 2)(x + 2020) > 0 `

   ` => ` \(\left[ \begin{array}{l}\left \{ {{x-2>0} \atop {x+2020>0}} \right. \\\left \{ {{x-2<0} \atop {x+2020<0}} \right. \end{array} \right.\) 

   ` => ` \(\left[ \begin{array}{l}\left \{ {{x>2} \atop {x>-2020}} \right. \\\left \{ {{x<2} \atop {x<-2020}} \right. \end{array} \right.\)

   ` => ` \(\left[ \begin{array}{l}x>2\\x<-2020\end{array} \right.\) 

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