Đáp án:
\(\begin{array}{l}
B6)\\
a)Min = \dfrac{{19}}{4}\\
b)Min = 100\\
B7)\\
a)Max = 36\\
b)Max = 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B6)\\
a)A = {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{19}}{4}\\
= {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{19}}{4}\\
Do:{\left( {x + \dfrac{3}{2}} \right)^2} \ge 0\forall x\\
\to {\left( {x + \dfrac{3}{2}} \right)^2} + \dfrac{{19}}{4} \ge \dfrac{{19}}{4}\\
\to Min = \dfrac{{19}}{4}\\
\Leftrightarrow x = - \dfrac{3}{2}\\
b)B = \left( {{x^2} - 7x + 10} \right)\left( {{x^2} - 7x - 10} \right)\\
= {\left( {{x^2} - 7x} \right)^2} - 100\\
Do:{\left( {{x^2} - 7x} \right)^2} \ge 0\forall x\\
\to {\left( {{x^2} - 7x} \right)^2} - 100 \ge 100\\
\to Min = 100\\
\Leftrightarrow {\left( {{x^2} - 7x} \right)^2} = 0\\
\to {x^2} - 7x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 7
\end{array} \right.\\
B7)\\
a)A = - \left( {{x^2} + 10x - 11} \right)\\
= - \left( {{x^2} + 10x + 25 - 36} \right)\\
= - {\left( {x + 5} \right)^2} + 36\\
Do:{\left( {x + 5} \right)^2} \ge 0\forall x\\
\to - {\left( {x + 5} \right)^2} \le 0\\
\to - {\left( {x + 5} \right)^2} + 36 \le 36\\
\to Max = 36\\
\Leftrightarrow x = - 5\\
b)B = 2\left| {x - 4} \right| - {\left| {x - 4} \right|^2}\\
= - \left( {{{\left| {x - 4} \right|}^2} - 2\left| {x - 4} \right|} \right)\\
= - \left( {{{\left| {x - 4} \right|}^2} - 2\left| {x - 4} \right| + 1 - 1} \right)\\
= - {\left( {\left| {x - 4} \right| - 1} \right)^2} + 1\\
Do:{\left( {\left| {x - 4} \right| - 1} \right)^2} \ge 0\forall x\\
\to - {\left( {\left| {x - 4} \right| - 1} \right)^2} \le 0\\
\to - {\left( {\left| {x - 4} \right| - 1} \right)^2} + 1 \le 1\\
\to Max = 1\\
\Leftrightarrow \left| {x - 4} \right| - 1 = 0\\
\to \left| {x - 4} \right| = 1\\
\to \left[ \begin{array}{l}
x - 4 = 1\\
x - 4 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.
\end{array}\)
