\(3.A=\dfrac{2\sqrt{x}+17}{\sqrt{x}+5}=\dfrac{2\left(\sqrt{x}+5\right)+7}{\sqrt{x}+5}\)\(=2+\dfrac{7}{\sqrt{x}+5}\)
\(\sqrt{x}+5\ge5=>2+\dfrac{7}{\sqrt{x}+5}\le2+\dfrac{7}{5}=3,4\)
dấu'=' xảy ra<=>x=0=>MaxA=3,4
3.\(A=\dfrac{2\sqrt{x}+17}{\sqrt{x}+5}=2+\dfrac{7}{\sqrt{x}+5}\)
\(A\in Z< =>\sqrt{x}+5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(=>x\in\left\{4\right\}\)