`B = x^2 + 4x + 5`

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

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

Vì `(x + 2)^2 \ge 0 \forall x` nên `(x + 2)^2 + 1 \ge 1`

Dấu "`=`" xảy ra khi `x + 2 = 0 \Rightarrow x = -2`

Vậy `B_{min} = 1` khi `x = -2`

`C = x^2 + 6x - 3`

`= (x^2 + 6x + 9) - 12`

`= (x + 3)^2 - 12`

Vì `(x + 3)^2 \ge 0 \forall x` nên `(x + 3)^2 - 12 \ge -12`

Dấu "`=`" xảy ra khi `x + 3 = 0 \Rightarrow x = -3`

Vậy `C_{min} = -12` khi `x = -3`

`D = x^2 + 3x + 1`

`= (x^2 + 2x * 3/2 + 9/4) - 5/4`

`= (x + 3/2)^2 - 5/4`

Vì `(x + 3/2)^2 \ge 0 \forall x` nên `(x + 3/2)^2 - 5/4 \ge -5/4`

Dấu "`=`" xảy ra khi `x + 3/2 = 0 \Rightarrow x = -3/2`

Vậy `D_{min} = -5/4` khi `x = -3/2`