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

Toán Lớp 7: anh em làm hộ mình phát: a,(5-x).(2+x)<0 b,(5-x).(x+2)>o

Toán Lớp 7: anh em làm hộ mình phát:
a,(5-x).(2+x)<0 b,(5-x).(x+2)>o

Comments ( 2 )

  1. ~rai~
    \(a)(5-x)(2+x)<0\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}5-x>0\\2+x<0\end{cases}\\\begin{cases}5-x<0\\2+x>0\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}x<5\\x<-2\end{cases}\\\begin{cases}x>5\\x>-2\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x<-2\\x>5.\end{array}\right.\\b)(5-x)(x+2)>0\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}5-x>0\\x+2>0\end{cases}\\\begin{cases}5-x<0\\x+2<0\end{cases}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}\begin{cases}x<5\\x>-2\end{cases}\\\begin{cases}x>5\\x<-2\end{cases}\quad\text{(vô lí)}\end{array}\right.\\\Leftrightarrow -2<x<5.\)
     

  2. $a) (5-x).(2+x)<0$
    TH $1$ : \(\left[ \begin{array}{l}5-x<0\\2+x>0\end{array} \right.\) =>\(\left[ \begin{array}{l}x>5\\x>-2\end{array} \right.\) $=>x>5$
    TH $2$: \(\left[ \begin{array}{l}5-x>0\\2+x<0\end{array} \right.\) =>\(\left[ \begin{array}{l}x<5\\x<-2\end{array} \right.\) =>$x<-2$
    Vậy với mọi $5<x<-2$ (vô lí) 
    Phương trình kh thể xác định nghiệm 
    $b) ,(5-x).(x+2)>0
    TH $1$ : \(\left[ \begin{array}{l}5-x<0\\2+x<0\end{array} \right.\) =>\(\left[ \begin{array}{l}x>5\\x<-2\end{array} \right.\) (vô lí)
    TH $2$: \(\left[ \begin{array}{l}5-x>0\\2+x>0\end{array} \right.\) =>\(\left[ \begin{array}{l}x<5\\x>-2\end{array} \right.\) =>$-2<x<5$
    Vậy với $-2<x<5$ thì $(5-x).(x+2)>0$
     

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