a) Ta có: \(A=\left(\sqrt{2}-8\sqrt{32}+2\sqrt{450}\right):\left(-3\sqrt{8}\right)\)
\(=\left(\sqrt{2}-32\sqrt{2}+30\sqrt{2}\right):\left(-6\sqrt{2}\right)\)
\(=\dfrac{-\sqrt{2}}{-6\sqrt{2}}=\dfrac{1}{6}\)
Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}-\dfrac{6}{9-x}\right):\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{x+5\sqrt{x}+6+\sqrt{x}-3+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{1}\)
\(=\dfrac{x+6\sqrt{x}+9}{\sqrt{x}+3}\)
\(=\sqrt{x}+3\)
b) Để \(AB\le1\) thì \(AB-1\le0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+3}{6}-\dfrac{6}{6}\le0\)
\(\Leftrightarrow\sqrt{x}\le3\)
hay \(x\le9\)
Kết hợp ĐKXĐ, ta được: \(0\le x< 9\)
