Đáp án:
Giải thích các bước giải:
`38-|x-10|=(-6)^20 : (9^9 . 4^10)`
`=>38-|x-10|=6^20 : ((3^2)^9 . (2^2)^10)`
`=>38-|x-10|=(2.3)^20 :(3^18 . 2^20)`
`=>38-|x-10|=2^20. 3^20: 3^18: 2^20`
`=>38-|x-10|=(2^20:2^20).(3^20:3^18)`
`=>38-|x-10|=1.3^2`
`=>38-|x-10|=9`
`=>|x-10|=29`
`=>` \(\left[ \begin{array}{l}x-10=29\\x-10=-29\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=29+10\\x=-29+10\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=39\\x=-19\end{array} \right.\)
Vậy `x \in {39;-19}`