a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
\(Q=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{14}{9-x}\right)\cdot\dfrac{\sqrt{x}-3}{2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)^2+\left(\sqrt{x}+3\right)^2+14}{\left(x-9\right)}\cdot\dfrac{\sqrt[]{x}-3}{2}\)
\(=\dfrac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\sqrt{x}+3}\cdot\dfrac{1}{2}\)
\(=\dfrac{2x+32}{2\left(\sqrt{x}+3\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
b: \(Q=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\)
\(=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6>=2\cdot\sqrt{\left(\sqrt[]{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6\)
=>\(Q>=2\cdot5-6=10-6=4\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}+3=\sqrt{25}=5\)
=>x=4(nhận)
