a: \(Q=\dfrac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{x-9}\cdot\dfrac{\sqrt{x}-3}{2}\)
\(=\dfrac{2x+32}{2}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{x+16}{\sqrt{x}+3}\)
b: \(Q=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)
=>\(Q=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\)
Dấu = xảy ra khi căn x+3=5
=>x=4