Đáp án:
\(\begin{array}{l}
1)Min = 24\\
2)Min = \dfrac{{59}}{{12}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)A = {x^2} - 6x - 15\\
= {x^2} - 6x + 9 - 24\\
= {\left( {x - 3} \right)^2} - 24\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to {\left( {x - 3} \right)^2} - 24 \ge - 24\\
\to Min = 24\\
\Leftrightarrow x - 3 = 0\\
\to x = 3\\
2)B = 3{x^2} - 11x + 15\\
= 3{x^2} - 2.x\sqrt 3 .\dfrac{{11}}{{2\sqrt 3 }} + \dfrac{{121}}{{12}} + \dfrac{{59}}{{12}}\\
= {\left( {x\sqrt 3 - \dfrac{{11}}{{2\sqrt 3 }}} \right)^2} + \dfrac{{59}}{{12}}\\
Do:{\left( {x\sqrt 3 - \dfrac{{11}}{{2\sqrt 3 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 3 - \dfrac{{11}}{{2\sqrt 3 }}} \right)^2} + \dfrac{{59}}{{12}} \ge \dfrac{{59}}{{12}}\\
\to Min = \dfrac{{59}}{{12}}\\
\Leftrightarrow x\sqrt 3 - \dfrac{{11}}{{2\sqrt 3 }} = 0\\
\to x = \dfrac{{11}}{6}
\end{array}\)
