`1)` $5^{2x-3}$ `- 2` . $5^{2}$ `=` $5^{2}$ . `3`
$5^{2x}$ `-` $5^{3}$ `-` `2` `.` `25` `=` `25.3`
$25^{x}$ `-` `125` `-` `50` `=` `75`
$25^{x}$ `-` `(125+50)` `=` `75`
$25^{x}$ `-` `175` `=` `75`
$25^{x}$ `=` `75+175`
$25^{x}$ `=` `250`
`=>` `x=∅`
Vậy không có `x` thỏa mãn đề bài
`2)` Để `34x` $\vdots$ `5`
`=>` \(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\) `
Mà `34x` $\vdots$ `3`
`=>` `(3+4+x)` $\vdots$ `3` hay `(7+x)` $\vdots$ `3`
TH`1`: `x=0`
`=>` `7+0=7` $\not\vdots$ `3` `(`loại`)`
TH`2`: `x=5`
`=>` `7+5=12` $\vdots$ `3` `(`thỏa mãn`)`
Vậy `x=5`
