\(\dfrac{x+1}{16}-\dfrac{2x-1}{8}=\dfrac{-x-8}{2}\)
\(\Rightarrow\dfrac{x+1}{16}-\dfrac{4x-2}{16}+\dfrac{8\left(x+8\right)}{16}=0\)
\(\Rightarrow x+1-4x+2+8x+64=0\)
\(\Rightarrow5x+67=0\)
\(\Rightarrow5x=-67\)
\(\Rightarrow x=-\dfrac{67}{5}\)
`(x+1)/16-(2x-1)/8=(-x-8)/2`
`=>x+1-2(2x-1)=8(-x-8)`
`=>x+1-4x+2=-8x-64`
`=>-3x+3=-8x-64`
`=>5x=67`
`=>x=67/5`
Vậy `x=67/5`