a) Ta có: \(-\left|x+\dfrac{9}{13}\right|\le0\forall x\)
\(\Rightarrow2-\left|x+\dfrac{9}{13}\right|\le2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{-9}{13}\)
Vậy \(MAX\) \(A=2\Leftrightarrow x=\dfrac{-9}{13}\)
b) Ta có: \(-\left|\dfrac{3}{5}-x\right|\le0\forall x\)
\(\Rightarrow7-\left|\dfrac{3}{5}-x\right|\le7\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{5}\)
Vậy \(MAX\) \(B=7\Leftrightarrow x=\dfrac{3}{5}\)