\(A=\dfrac{3\left|x\right|+2}{4\left|x\right|-5}\)

\(\Leftrightarrow4A=\dfrac{12\left|x\right|+8}{4\left|x\right|-5}\)

\(\Leftrightarrow4A=\dfrac{3\left(4\left|x\right|-5\right)+23}{4\left|x\right|-5}\)

\(\Leftrightarrow4A=\dfrac{3\left(4\left|x\right|-5\right)}{4\left|x\right|-5}+\dfrac{23}{4\left|x\right|-5}\)

\(\Leftrightarrow4A=3+\dfrac{23}{4\left|x\right|-5}\)

We have \(\left|x\right|\ge0\Rightarrow4\left|x\right|\ge0\Rightarrow4\left|x\right|-5\ge-5\)

\(\Rightarrow\dfrac{1}{4\left|x\right|-5}\le-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{23}{4\left|x\right|-5}\le-\dfrac{23}{5}\)

\(\Rightarrow3+\dfrac{23}{4\left|x\right|-5}\le3+\left(-\dfrac{23}{5}\right)\)

\(\Rightarrow4A\le-\dfrac{8}{5}\)

\(\Rightarrow A\le-\dfrac{2}{5}\)

\(\Rightarrow MaxA=-\dfrac{2}{5}\) when \(x=0\)

I don't know

But Ngân Hà is doing correct

Thank you