Đáp án:
Giải thích các bước giải:
`(x-2).(x+15)=0`
`⇔`\(\left[ \begin{array}{l}x-2 = 0\\x + 15 = 0 \end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = 2\\x = -15 \end{array} \right.\)
`⇒ x ∈ {2; -15}`
`(x-2).(x+4)=0`
`⇔` \(\left[ \begin{array}{l}x-2 = 0\\x + 4 = 0 \end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = 2\\x = -4 \end{array} \right.\)
`⇒ x ∈ {2 ; -4}`
`(7-x).(x+19)=0`
`⇔` \(\left[ \begin{array}{l}7 - x = 0\\x + 19 = 0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=7\\x=-19\end{array} \right.\)
`⇒ x ∈ {7; -19}`
`-5 < x <1`
Mà `-5 < -4; -3; -2; -1; 0 < 1`
`⇒ x ∈ {-4; -3; -2; -1; 0}`
`|x|< 3`
`⇔ |x| ∈ {1; 2; 0}`
`⇒ x ∈ {±1; ± 2; 0}`
