*)max:Ta có:0$\leq$xy$\leq$$\frac{(x+y)^{2}}{4}$=$\frac{1}{4}$$\Rightarrow$$\frac{-1}{16}$$\leq$xy-$\frac{1}{16}$$\leq$$\frac{3}{16}$$\Rightarrow$$\left| {xy-\frac{1}{16}} \right|$$\leq$$\frac{3}{16}$S=16$(xy-\frac{1}{16})^{2}$+$\frac{191}{16}$$\leq$$\frac{25}{2}$Dấu ''='' xra $\Leftrightarrow$x=y=$\frac{1}{2}$
*)max:Ta có:0$\leq$xy$\leq$$\frac{(x+y)^{2}}{4}$=$\frac{1}{4}$$\Rightarrow$$\frac{-1}{16}$$\leq$xy-$\frac{1}{16}$$\leq$$\frac{3}{16}$$\Rightarrow$$\left| {xy-\frac{1}{16}} \right|$$\leq$$\frac{3}{16}$S=16$(xy-\frac{1}{16})^{2}$+$\frac{191}{16}$$\leq$$\frac{25}{2}$Dấu ''='' xra $\Leftrightarrow$x=x=$\frac{1}{2}$
*)max:Ta có:0$\leq$xy$\leq$$\frac{(x+y)^{2}}{4}$=$\frac{1}{4}$$\Rightarrow$$\frac{-1}{16}$$\leq$xy-$\frac{1}{16}$$\leq$$\frac{3}{16}$$\Rightarrow$$\left| {xy-\frac{1}{16}} \right|$$\leq$$\frac{3}{16}$S=16$(xy-\frac{1}{16})^{2}$+$\frac{191}{16}$$\leq$$\frac{25}{2}$Dấu ''='' xra $\Leftrightarrow$x=
y=$\frac{1}{2}$