`\color{#a8b0b4}{⸻}\color{#91999d}{⸻}\color{#7a8286}{『}\color{#636b6f}{♡} \color{#4c5356}{E}\color{#434a4d}{m}\color{#3b4244}{p}\color{#2a3133}{t}\color{#21282a}{y}\color{#1a2123}{B}\color{#141a1c}{o} \color{#0f1516}{y} \color{#05090a}{♡}\color{#0a1011}{』}\color{#141a1c}{⸻}\color{#1a2123}{⸻}`

`d)`

`D=-x^2+6x-11`

`D=-(x^2-6x+9-9)-11`

`D=-[(x-3)^2-9]-11`

`D=-(x-3)^2+9-11`

`D=-(x-3)^2-2`

Vì `-(x-3)^2<=0\forallx` nên `D<=-2`

Vậy `D_(max)=-2<=>x=3`

`e)`

`E=5-8x-x^2`

`=-(x^2+8x+16-16)+5`

`=-[(x+4)^2-16]+5`

`=-(x+4)^2+21`

Vì `-(x+4)^2<=0\forallx` nên `E<=21`

Vậy `E_(max)=21<=>x=-4`

`f)`

`F=-(x^2-4x)+1`

`=-(x^2-4x+4-4)+1`

`=-[(x-2)^2-4]+1`

`=-(x-2)^2+1+4`

`=-(x-2)^2+5`

Vì `-(x-2)^2<=0\forallx` nên `F<=5`

Vậy `F_(max)=5<=>x=2`