[tex]-x^{2}-4xy-6y^{2}+x-8y-2017=-(x^{2}+4xy+4y^{2})+(x+2y)-(2y^{2}+10y)-2017=-(x+2y)^{2}+2.(x+2y).\frac{1}{2}-\frac{1}{4}-2(y^{2}+2.y.\frac{5}{2}+\frac{25}{4})+\frac{1}{4}+\frac{25}{2}-2017=-(x+2y-\frac{1}{2})^{2}-2(y+\frac{5}{2})^{2}-2004,25\leq-2004,25 [/tex]
Dấu "=" có khi [tex]y=\frac{-5}{2}[/tex] và [tex]x=\frac{11}{2}[/tex]