`1)|x|=-5`
Vì `|x|≥0` nên không tồn tại `x`.
Vậy `x in∅`
`2)|x|=7`
\(\left[ \begin{array}{l}x=7\\x=-7\end{array} \right.\)
Vậy `x in{±7}`
`3)|-x|=-9`
`|x|=-9`
Vì `|x|≥0` nên không tồn tại `x`
Vậy `x in∅`
`4)-|x|=11`
Vì `-|x|≤0` nên không tồn tại `x`
Vậy `x in∅`
`5)-|x|=-24`
`|x|=24`
\(\left[ \begin{array}{l}x=24\\x=-24\end{array} \right.\)
Vậy `x in{±24}`
`25)|2x-5|=7`
\(\left[ \begin{array}{l}2x-5=7\\2x-5=-7\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
Vậy `x in{6;-1}`
`26)|3x-7|=|-20|`
`|3x-7|=20`
\(\left[ \begin{array}{l}3x-7=20\\3x-7=-20\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=9\\x=\dfrac{13}{3}\end{array} \right.\)
Vậy `x in{9;13/3}`
`27)36-|2x-1|=15`
`|2x-1|=36-15`
`|2x-1|=21`
`\(\left[ \begin{array}{l}2x-1=21\\2x-1=-21\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=11\\x=-10\end{array} \right.\)
Vậy `x in{11;-10}`
`28)|4x-9|(-8)=-88`
`|4x-9|=-88:(-8)`
`|4x-9|=11`
\(\left[ \begin{array}{l}4x-9=11\\4x-9=-11\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac{19}{4}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `x in{19/4;-1/2}`
`29)|5x-11|(-8)=-|-32|`
`|5x-11|(-8)=-32`
`|5x-11|=-32:(-8)`
`|5x-11|=4`
\(\left[ \begin{array}{l}5x-11=4\\5x-11=-4\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=3\\x=\dfrac{7}{5}\end{array} \right.\)
Vậy `x in{3;7/5}`