\(a.Q=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{14}{9-x}\right).\dfrac{\sqrt{x}-3}{2}=\dfrac{\left(\sqrt{x}-3\right)^2+\left(\sqrt{x}+3\right)^2+14}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{2}=\dfrac{2\left(x+16\right)}{2\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\left(x\ge0;x\ne9\right)\)
\(b.Q=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\sqrt{x}+3}+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}=2.5=10\)
\(\Leftrightarrow\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge10-6=4\)
Dấu \("="\) xảy ra khi \(x=4\left(TM\right)\)
\(\Rightarrow Q_{Min}=4."="\Leftrightarrow x=4\)
