Đáp án:
f. \(Min = \dfrac{{47}}{{12}}\)
Giải thích các bước giải:
\(\begin{array}{l}
b.B = {x^2} + 7x - 1\\
= {x^2} + 2x.\dfrac{7}{2} + \dfrac{{49}}{4} - \dfrac{{53}}{4}\\
= {\left( {x + \dfrac{7}{2}} \right)^2} - \dfrac{{53}}{4}\\
Do:{\left( {x + \dfrac{7}{2}} \right)^2} \ge 0\\
\to {\left( {x + \dfrac{7}{2}} \right)^2} - \dfrac{{53}}{4} \ge - \dfrac{{53}}{4}\\
\to Min = - \dfrac{{53}}{4}\\
\Leftrightarrow x + \dfrac{7}{2} = 0\\
\Leftrightarrow x = - \dfrac{7}{2}\\
d.D = {x^2} + {\left( { - y} \right)^2} + 1 - 2xy - 2y + 2x + {x^2} - 2x + 1 + 1998\\
= {\left( {x - y + 1} \right)^2} + {\left( {x - 1} \right)^2} + 1998\\
Do:\left\{ \begin{array}{l}
{\left( {x - y + 1} \right)^2} \ge 0\forall x;y \in R\\
{\left( {x - 1} \right)^2} \ge 0\forall x;y \in R
\end{array} \right.\\
\to {\left( {x - y + 1} \right)^2} + {\left( {x - 1} \right)^2} \ge 0\\
\to {\left( {x - y + 1} \right)^2} + {\left( {x - 1} \right)^2} + 1998 \ge 1998\\
\to Min = 1998\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y + 1 = 0\\
x = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
e.E = {x^2} + {\left( { - y} \right)^2} + 9 - 2xy + 2.3.x - 2.3.y + 4{y^2} - 2.2y.3 + 9 + 32\\
= {\left( {x - y + 3} \right)^2} + {\left( {2y - 3} \right)^2} + 32\\
Do:\left\{ \begin{array}{l}
{\left( {x - y + 3} \right)^2} \ge 0\\
{\left( {2y - 3} \right)^2} \ge 0
\end{array} \right.\forall x;y \in R\\
\to {\left( {x - y + 3} \right)^2} + {\left( {2y - 3} \right)^2} \ge 0\\
\to {\left( {x - y + 3} \right)^2} + {\left( {2y - 3} \right)^2} + 32 \ge 32\\
\to Min = 32\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y + 3 = 0\\
2y - 3 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{3}{2}\\
x = - \dfrac{3}{2}
\end{array} \right.\\
f.F = {\left( {x\sqrt 3 } \right)^2} + 2.x\sqrt 3 .\dfrac{1}{{2\sqrt 3 }} + \dfrac{1}{{12}} + \dfrac{{47}}{{12}}\\
= {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{47}}{{12}}\\
Do:{\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }}} \right)^2} + \dfrac{{47}}{{12}} \ge \dfrac{{47}}{{12}}\\
\to Min = \dfrac{{47}}{{12}}\\
\Leftrightarrow x\sqrt 3 + \dfrac{1}{{2\sqrt 3 }} = 0\\
\Leftrightarrow x = - \dfrac{1}{6}
\end{array}\)
