Đáp án:
MinD=-36
Giải thích các bước giải:
\(\begin{array}{l}
A = {x^2} + 2x - 3\\
= {x^2} + 2x + 1 - 4\\
= {\left( {x + 1} \right)^2} - 4\\
Do:{\left( {x + 1} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x + 1} \right)^2} - 4 \ge - 4\\
\to Min = - 4\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\\
b)B = {x^2} - 4x + 1 = {x^2} - 4x + 4 - 3\\
= {\left( {x - 2} \right)^2} - 3\\
Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to {\left( {x - 2} \right)^2} - 3 \ge - 3\\
\to Min = - 3\\
\Leftrightarrow x = 2\\
C = 4{x^2} + 4x + 11 = 4{x^2} + 4x + 1 + 10\\
= {\left( {2x + 1} \right)^2} + 10\\
Do:{\left( {2x + 1} \right)^2} \ge 0\forall x\\
\to {\left( {2x + 1} \right)^2} + 10 \ge 10\\
\to Min = 10\\
\Leftrightarrow x = - \dfrac{1}{2}\\
D = \left( {x - 1} \right)\left( {x + 6} \right)\left( {x + 3} \right)\left( {x + 2} \right)\\
= \left( {{x^2} + 5x - 6} \right)\left( {{x^2} + 5x + 6} \right)\\
= {\left( {{x^2} + 5x} \right)^2} - 36\\
Do:{\left( {{x^2} + 5x} \right)^2} \ge 0\forall x\\
\to {\left( {{x^2} + 5x} \right)^2} - 36 \ge - 36\\
\to Min = - 36\\
\Leftrightarrow {x^2} + 5x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.
\end{array}\)
