Đáp án:
B13:
b) Min=10
Giải thích các bước giải:
\(\begin{array}{l}
B12:\\
A = {x^2} - 6x + 10\\
= {x^2} - 6x + 9 + 1\\
= {\left( {x - 3} \right)^2} + 1\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to {\left( {x - 3} \right)^2} + 1 > 0\\
\to A > 0\forall x\\
B = {x^2} - 2x + 1 + 9{y^2} - 6y + 1 + 1\\
= {\left( {x - 1} \right)^2} + {\left( {3y - 1} \right)^2} + 1\\
Do:{\left( {x - 1} \right)^2} + {\left( {3y - 1} \right)^2} \ge 0\forall x;y\\
\to {\left( {x - 1} \right)^2} + {\left( {3y - 1} \right)^2} + 1 > 0\\
\to B > 0\forall x;y\\
B13:\\
A = {x^2} - 4x + 4 - 3\\
= {\left( {x - 2} \right)^2} - 3\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to {\left( {x - 2} \right)^2} - 3 \ge - 3\\
\to Min = - 3\\
\Leftrightarrow x - 2 = 0\\
\Leftrightarrow x = 2\\
B = 4{x^2} + 4x + 1 + 10\\
= {\left( {2x + 1} \right)^2} + 10\\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {2x + 1} \right)^2} + 10 \ge 10\\
\to Min = 10\\
\Leftrightarrow x = - \dfrac{1}{2}
\end{array}\)
