`#tnvt`
`1)`
`125-8x^3`
`=5^3-(2x)^3`
`=(5-2x)(25+10x+4x^2)`
`2)`
`a)(4x-1)(2x+3)-8x^2=5`
`<=>8x^2+12x-2x-3-8x^2=5`
`<=>10x=8`
`<=>x=4/5`
Vậy `x=4/5`
`b)16-(2x-1)^2=0`
`<=>4^2-(2x-1)^2=0`
`<=>(4-2x+1)(4+2x-1)=0`
`<=>(5-2x)(3+2x)=0`
`<=>[(5-2x=0),(3+2x=0):}`
`<=>`$\left[\begin{matrix} x=\dfrac{5}{2} \\ x=\dfrac{-3}{2} \end{matrix}\right.$
Vậy `x\in{5/2;-3/2}`