tìm max của biểu thức: a) A=129-5|x-7| b) B= x+1/2 -|x-2/3|

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
A = 129 – 5\left| {x – 7} \right|\\
\left| {x – 7} \right| \ge 0,\forall x \Rightarrow 5\left| {x – 7} \right| \ge 0 \Rightarrow A = 129 – 5\left| {x – 7} \right| \le 129\\
{A_{\max }} = 129 \Leftrightarrow x = 7\\
b,\\
B = x + \frac{1}{2} – \left| {x – \frac{2}{3}} \right|\\
\left| a \right| \ge a,\,\,\,\,\forall a \Rightarrow \left| {x – \frac{2}{3}} \right| \ge x – \frac{2}{3}\\
 \Rightarrow B = x + \frac{1}{2} – \left| {x – \frac{2}{3}} \right| \le x + \frac{1}{2} – \left( {x – \frac{2}{3}} \right) = \frac{7}{6}\\
 \Rightarrow {B_{\max }} = \frac{7}{6} \Leftrightarrow x = \frac{2}{3}\\
d,\\
D = \frac{{2004}}{{2003}} – \left| {x – \frac{3}{5}} \right|\\
\left| {x – \frac{3}{5}} \right| \ge 0 \Rightarrow D = \frac{{2004}}{{2003}} – \left| {x – \frac{3}{5}} \right| \le \frac{{2004}}{{2003}}\\
 \Rightarrow {D_{\max }} = \frac{{2004}}{{2003}} \Leftrightarrow x = \frac{3}{5}
\end{array}\)