Đk:\(a>2\)
\(\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
Đặt \(b=\sqrt{a-2}\Leftrightarrow a=b^2+2\)
Biểu thức \(\Leftrightarrow\dfrac{b+2}{3}\left(\dfrac{b}{3+b}+\dfrac{b^2+2+7}{11-b^2-2}\right):\left(\dfrac{3b+1}{b^2-3b}-\dfrac{1}{b}\right)\)
\(=\dfrac{b+2}{3}\left[\dfrac{b}{3+b}-\dfrac{b^2+9}{b^2-9}\right]:\left[\dfrac{3b+1}{b\left(b-3\right)}-\dfrac{b-3}{b\left(b-3\right)}\right]\)
\(=\dfrac{b+2}{3}.\dfrac{b\left(b-3\right)-b^2-9}{\left(b-3\right)\left(3+b\right)}:\dfrac{3b+1-\left(b-3\right)}{b\left(b-3\right)}\)
\(=\dfrac{b+2}{3}.\dfrac{-3\left(b+3\right)}{\left(b-3\right)\left(3+b\right)}.\dfrac{b\left(b-3\right)}{2\left(b+2\right)}\)
\(=-\dfrac{b}{2}\)
\(=\dfrac{\sqrt{a-2}}{-2}\)