Toán Lớp 6: Tìm số nguyên x biết
a)(x+1).(x-2)<0
b)(x+2).(x+6)>0
c)(x+1).(8-x)<0
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Toán Lớp 6: Tìm số nguyên x biết
a)(x+1).(x-2)<0
b)(x+2).(x+6)>0
c)(x+1).(8-x)<0
Comments ( 2 )
⇔$\left \{ {{x>-1} \atop {x<2}} \right.$
⇔-1<x<2
⇔$\left \{ {{x<-1} \atop {x>2}} \right.$
⇔2<x<-1 (vô lý)
⇔\(\left[ \begin{array}{l}\left \{ {{x>-2} \atop {x>-6}} \right. \\\left \{ {{x<-2} \atop {x<-6}} \right. \end{array} \right.\)
⇔\(\left[ \begin{array}{l}x>-2\\x<-6\end{array} \right.\)
Vậy S={x∈Z l x>-2 hoặc x<-6}
⇔\(\left[ \begin{array}{l}\left \{ {{x>-1} \atop {-x<-8}} \right. \\\left \{ {{x<-1} \atop {-x>-8}} \right. \end{array} \right.\)
⇔\(\left[ \begin{array}{l}\left \{ {{x>-1} \atop {x>8}} \right. \\\left \{ {{x<-1} \atop {x<8}} \right. \end{array} \right.\)
⇔\(\left[ \begin{array}{l}x>8\\x<-1\end{array} \right.\)
Vậy S={x∈Z l x>8 hoặc x<-1}