`a) 12/(3+|5x+1|+|2y-1|)`
Do `|5x+1|>=0;|2y-1|>=0` với `AAx;y`
`=> |5x+1|+|2y-1|+3>=3`
`=> 12/(3+|5x+1|+|2y-1|)<=12/3=4`
Dấu = xảy ra khi `{(5x+1=0),(2y-1=0):}<=>`$\begin{cases}x=\dfrac{-1}{5}\\y=\dfrac{1}{2}\end{cases}$
Vậy `max=4<=>(x;y)=(-1/5;1/2)`
`b) 5/(4x^2+4x+2y+y^2+3)`
Có `4x^2+4x+2y+y^2+3`
`=(4x^2+4x+1)+(y^2+2y+1)+1`
`=(2x+1)^2+(y+1)^2+1>=1`
`=> 5/(4x^2+4x+2y+y^2+3)<=5`
Dấu = xảy ra khi `{(2x+1=0),(y+1=0):}<=>`$\begin{cases}x=\dfrac{-1}{2}\\y=-1\end{cases}$
Vậy `max=5<=>(x;y)=(-1/2;-1)`
