Đáp án:

c. \(Min = \dfrac{3}{4}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.A = {y^2} + 2.4.y + 16 - 1\\
 = {\left( {y + 4} \right)^2} - 1\\
Do:{\left( {y + 4} \right)^2} \ge 0\forall y\\
 \to {\left( {y + 4} \right)^2} - 1 \ge  - 1\\
 \to Min =  - 1\\
 \Leftrightarrow y =  - 4\\
b.B = {x^2} + {y^2} - 2x + 2y + 15\\
 = {x^2} - 2x + 1 + {y^2} + 2y + 1 + 13\\
 = {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} + 13\\
Do:\left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0\\
{\left( {y + 1} \right)^2} \ge 0
\end{array} \right.\forall x;y \in R\\
 \to {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} \ge 0\\
 \to {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} + 13 \ge 13\\
 \to Min = 13\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y =  - 1
\end{array} \right.\\
c.C = {x^2} + {y^2} - x + 6y + 10\\
 = {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} + {y^2} + 2.3.y + 9 + \dfrac{3}{4}\\
 = {\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 3} \right)^2} + \dfrac{3}{4}\\
Do:\left\{ \begin{array}{l}
{\left( {x - \dfrac{1}{2}} \right)^2} \ge 0\\
{\left( {y + 3} \right)^2} \ge 0
\end{array} \right.\forall x;y\\
 \to {\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 3} \right)^2} \ge 0\\
 \to {\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 3} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
 \to Min = \dfrac{3}{4}\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y =  - 3
\end{array} \right.
\end{array}\)

$a) A=y^2+8y+15$

$=y^2+8y+16-1$

$=(y+4)^2-1$

Vì $(y+4)^2 \geq 0 ∀y$

$⇒(y+4)^2-1 \geq -1 ∀y$

$⇒min_A=-1$ khi $y+4=0 ⇔ y=-4$

$b) B=x^2+y^2-2x+2y+15$

$=(x^2-2x+1)+(y^2+2y+1)+13$

$=(x-1)^2+(y+1)^2+13$

Vì $(x-1)^2+(y+1)^2 \geq 0 ∀x;y$

$⇒(x-1)^2+(y+1)^2+13 \geq 13 ∀x;y$

⇒ $min_B=13$ khi $\left\{ \begin{matrix}x=1\\y=-1\end{matrix} \right.$

$c) C=x^2+y^2-x+6y+10$

$=\left (x^2-x+\dfrac{1}{4} \right )+\dfrac{3}{4}+(y^2+6y+9)$

$=\left (x-\dfrac{1}{2} \right )+(y+3)^2+\dfrac{3}{4}$

$\left (x-\dfrac{1}{2} \right )+(y+3)^2 \geq 0 ∀x;y$

$⇒\left (x-\dfrac{1}{2} \right )+(y+3)^2+\dfrac{3}{4} \geq \dfrac{3}{4} ∀x;y$

$⇒min_C=\dfrac{3}{4}$ khi $\left\{ \begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix} \right.$