`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{1}{5}\right)^{2x-1}=\dfrac{1}{125}\)
`=>`\(\left(\dfrac{1}{5}\right)^{2x-1}=\left(\dfrac{1}{5}\right)^3\)
`=>`\(2x-1=3\)
`=> 2x = 3 + 1`
`=> 2x = 4`
`=> x = 4 \div 2`
`=> x = 2`
Vậy, `x = 2.`
`@` `\text {Kaizuu lv uuu}`