\(a,A=\dfrac{3+\sqrt{\dfrac{4}{9}}}{\sqrt{\dfrac{4}{9}}}=\dfrac{3+\dfrac{2}{3}}{\dfrac{2}{3}}=\dfrac{11}{3}:\dfrac{2}{3}=\dfrac{11}{2}\\ b,B=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\left(x>0;x\ne9\right)\\ B=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
a)Ta có \(x=\dfrac{4}{9}\Rightarrow\sqrt{x}=\dfrac{2}{3}\)
=>\(A=\dfrac{3+\sqrt{x}}{\sqrt{x}}=\dfrac{3+\dfrac{2}{3}}{\dfrac{2}{3}}=\dfrac{11}{2}\)
Vậy\(A=\dfrac{11}{2}khix=\dfrac{4}{9}\)
b)\(B=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
giải hộ mk vs chỉ cần có cách làm câu b thôi câu a thì ko cần
