Đáp án:
d) \(Max = - 2\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Do:{\left( {x - 2} \right)^2} \ge 0\forall x \in R\\
\to {\left( {x - 2} \right)^2} + 3 \ge 3\\
\to Min = 3\\
\Leftrightarrow x - 2 = 0\\
\to x = 2\\
b)Do:\left| {x - \dfrac{4}{7}} \right| \ge 0\forall x\\
\to \left| {x - \dfrac{4}{7}} \right| - \dfrac{1}{2} \ge - \dfrac{1}{2}\\
\to Min = - \dfrac{1}{2}\\
\Leftrightarrow x - \dfrac{4}{7} = 0\\
\Leftrightarrow x = \dfrac{4}{7}\\
c.Do:\left| {x - 3,5} \right| \ge 0\forall x\\
\to - \left| {x - 3,5} \right| \le 0\\
\to \dfrac{1}{2} - \left| {x - 3,5} \right| \le \dfrac{1}{2}\\
\to Max = \dfrac{1}{2}\\
\Leftrightarrow x = 3,5\\
d.Do:\left| {1,4 - x} \right| \ge 0\\
\to - \left| {1,4 - x} \right| \le 0\\
\to - \left| {1,4 - x} \right| - 2 \le - 2\\
\to Max = - 2\\
\Leftrightarrow x = 1,4
\end{array}\)
