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`5x=2y`
`->x/2=y/5` (1)
`2y=z+1`
`-> 2/(z+1)=1/y`
`-> (z+1)/2=y/1`
`-> y/1 . 1/5 =(z+1)/2 . 1/5`
`-> y/5=(z+1)/10` (2)
Từ (1), (2)
`-> x/2=y/5=(z+1)/10`
Đặt `x/2=y/5=(z+1)/10 =k (k \ne 0)`
`->` $\begin{cases} \dfrac{x}{2}=k\\\dfrac{y}{5}=k\\\dfrac{z+1}{10}=k \end{cases}$ `->` $\begin{cases} x=2k\\y=5k\\z=10k-1\end{cases}$
`A = (x+y-2z-2)/(x+5y+z+1)`
`-> A = (2k + 5k - 2 (10k-1) -2)/(2k + 5 . 5k + 10k - 1+1)`
`-> A = (2k + 5k - 20k + 2 - 2)/(2k + 25k + 10k - 1+1)`
`-> A = ( (2k+5k-20k) + (2-2) )/( (2k+25k + 10k) + (-1+1) )`
`-> A = (-13k)/(37k)`
`-> A=(-13)/37`
Vậy `A=(-13)/37` khi `5x=2y, 2y=z+1`
