$\begin{array}{l}+) \quad A = 3x^2 + 5x\\ \to A = 3\left(x^2 + 2.\dfrac{5}{6}x + \dfrac{25}{36}\right) - \dfrac{25}{12}\\ \to A = 3\left(x + \dfrac{5}{6}\right)^2 - \dfrac{25}{12}\\ Ta\,\,có:\\ \left(x + \dfrac{5}{6}\right)^2 \geq 0\,\,\forall x\\ \to 3\left(x + \dfrac{5}{6}\right)^2 - \dfrac{25}{12} \geq - \dfrac{25}{12}\\ \to A \geq - \dfrac{25}{12}\\ \text{Dấu = xảy ra }\,\Leftrightarrow x = -\dfrac{5}{6}\\ Vậy\,\,\min A = - \dfrac{25}{12} \Leftrightarrow x = - \dfrac{5}{6}\\+) \quad B = x^2 + 8x\\ \to B = x^2 + 2.4x + 16 - 16\\ \to B = (x + 4)^2 - 16\\ Ta\,\,có:\\ (x+4)^2 \geq 0\,\,\forall x\\ \to (x+4)^2 - 16 \geq -16\\ \to B \geq - 16\\ \text{Dấu = xảy ra }\,\Leftrightarrow x = - 4\\ Vậy\,\,\min B = - 16 \Leftrightarrow x = -4\\+) \quad C = 3x^2 + 5x - 2\\ \to C = 3\left(x + \dfrac{5}{6}\right)^2 - \dfrac{25}{12} - 2\\ \to C = 3\left(x + \dfrac{5}{6}\right)^2 - \dfrac{49}{12}\\ Ta\,\,có:\\ \left(x + \dfrac{5}{6}\right)^2 \geq 0\,\,\forall x\\ \to 3\left(x + \dfrac{5}{6}\right)^2 - \dfrac{49}{12} \geq - \dfrac{49}{12}\\ \to C \geq - \dfrac{49}{12}\\ \text{Dấu = xảy ra }\,\Leftrightarrow x = -\dfrac{5}{6}\\ Vậy\,\,\min C = - \dfrac{37}{12} \Leftrightarrow x = - \dfrac{5}{6}\\+) \quad D = 2x^2 + 7x\\ \to D = 2\left(x^2 + 2.\dfrac{7}{4}x + \dfrac{49}{16}\right) - \dfrac{49}{8}\\ \to D = 2\left(x + \dfrac{7}{4}\right)^2 - \dfrac{49}{8}\\ Ta\,\,có:\\ \left(x + \dfrac{7}{4}\right)^2 \geq 0\,\,\forall x \\ \to 2\left(x + \dfrac{7}{4}\right)^2 - \dfrac{49}{8}\geq -\dfrac{49}{8}\\ \to D \geq - \dfrac{49}{8}\\ \text{Dấu = xảy ra }\,\Leftrightarrow x = - \dfrac{7}{4}\\ Vậy\,\,\min D = \dfrac{49}{8}\Leftrightarrow x = - \dfrac{7}{4}\\+) \quad  E = x^2 - 4x + y^2 - 8y + 6\\ \to E = (x^2 - 2.2x + 4) + (y^2 - 2.4y + 16) - 14\\ \to E = (x-2)^2 + (y - 4)^2 - 14\\ Ta\,\,có:\\ \begin{cases}(x-2)^2 \geq 0\,\,\forall x\\(y-4)^2 \geq 0\,\,\forall y\end{cases}\\ \to (x-2)^2 + (y - 4)^2 - 14 \geq - 14\\ \text{Dấu = xảy ra }\,\Leftrightarrow \begin{cases}x = 2\\y = 4\end{cases}\\ Vậy\,\,\min E = -14 \Leftrightarrow (x;y) = (2;4) \end{array}$

`A = 3x^2 + 5x`

`= 3(x^2 + 2.(5)/(6)x + 25/36 - 25/36)`

`= 3(x + 5/6)^2 - 25/12`

Vì `3(x + 5/6)^2 >= 0` với `AA x in RR`

`=> A >= -25/12` với `AA x in RR`

Dấu "=" xảy ra:

`<=> x = -5/6`

Vậy `A_{min} = -25/12` khi `x = -5/6`

`B = x^2 + 8x`

`= x^2 + 2.4.x + 16 - 16`

`= (x + 4)^2 - 16`

Vì `(x + 4)^2 >= 0` với `AA x in RR`

`=> B >= -16` với `AA x in RR`

Dấu "=" xảy ra

`<=> x = -4`

Vậy `B_{min} = -16` khi `x = -4`

`C = 3x^2 + 5x - 2`

`= 3(x^2 + 2.(5)/(6)x + 25/36 - 25/36) - 2`

`= 3(x + 5/6)^2 - 49/12`

Vì `3(x + 5/6)^2 >= 0` với `AA x in RR`

`=> C >= -49/12` với `AA x in RR`

Dấu "=" xảy ra:

`<=> x = -5/6`

Vậy `C_{min} = -49/12` khi `x = -5/6`

`D = 2x^2 + 7x`

`= 2(x^2 + 2.(7)/(4)x + 49/16 - 49/16)`

`= 2(x + 7/4)^2 - 49/8`

Vì `2(x + 7/4)^2 >= 0` với `AA x in RR`

`=> D >= -49/8` với `AA x in RR`

Dấu "=" xảy ra:

`<=> x = -7/4`

Vậy `C_{min} = -49/8` khi `x = -7/4`

`E = x^2 - 4x + y^2 - 8y + 6`

`= x^2 - 2.2.x + 4 + y^2 - 2.4.y + 16 - 14`

`= (x - 2)^2 + (y - 4)^2 - 14`

Vì `(x - 2)^2 + (y - 4)^2 >= 0` với `AA x in RR`

`=> E >= -14` 

Dấu "=" xảy ra:

`<=>` \(\left\{ \begin{array}{l}x = 2\\y = 4\end{array} \right.\)

Vậy ...