Tìm số nguyên x sao cho: ( 2x + 5). ( x – 3) < 0 (2-x). (x+ 3) > 0 06/12/2021 Bởi Daisy Tìm số nguyên x sao cho: ( 2x + 5). ( x – 3) < 0 (2-x). (x+ 3) > 0
( 2x + 5 )( x – 3 ) < 0 2x+5<0 x−3<0 x<−52 x<3 =>x<−52 ( 2 – x )( x + 3 ) < 0 2−x<0 x+3<0 x>2 =>x<−3 Bình luận
( 2x + 5 )( x – 3 ) < 0 ⇔ \(\left[ \begin{array}{l}2x+5<0\\x-3<0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x< \frac{-5}{2} \\x<3\end{array} \right.\) ⇒ x < $\frac{-5}{2}$ ( 2 – x )( x + 3 ) < 0 ⇔ \(\left[ \begin{array}{l}2-x<0\\x+3<0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x>2\\x<-3\end{array} \right.\) ~CHÚC BẠN HỌC TỐT~ Bình luận
( 2x + 5 )( x – 3 ) < 0
2x+5<0
x−3<0
x<−52
x<3
=>x<−52
( 2 – x )( x + 3 ) < 0
2−x<0
x+3<0
x>2
=>x<−3
( 2x + 5 )( x – 3 ) < 0
⇔ \(\left[ \begin{array}{l}2x+5<0\\x-3<0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x< \frac{-5}{2} \\x<3\end{array} \right.\)
⇒ x < $\frac{-5}{2}$
( 2 – x )( x + 3 ) < 0
⇔ \(\left[ \begin{array}{l}2-x<0\\x+3<0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x>2\\x<-3\end{array} \right.\)
~CHÚC BẠN HỌC TỐT~