a) Khi \(x=\dfrac{4}{9}\) vào A ta có:
\(A=\dfrac{5-\dfrac{4}{9}}{\sqrt{\dfrac{4}{9}}}=\dfrac{41}{6}\)
b) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{6}{\sqrt{x}+3}-\dfrac{2x+18}{x-9}\)
ĐKXĐ: \(x>0;x\ne9\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)-6\left(\sqrt{x}-3\right)-\left(2x+18\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{x+3\sqrt{x}-6\sqrt{x}+18-2x-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
a: Khi x=4/9 thì \(A=\left(5-\dfrac{4}{9}\right):\dfrac{2}{3}=\dfrac{41}{9}\cdot\dfrac{3}{2}=\dfrac{41}{6}\)
b: \(B=\dfrac{x+3\sqrt{x}-6\sqrt{x}+18-2x-18}{x-9}\)
\(=\dfrac{-x-3\sqrt{x}}{x-9}=-\dfrac{\sqrt{x}}{\sqrt{x}-3}\)