Question

@Nguyễn Lê Phước Thịnh: NÉ HỘ

Answer 1

\(B=\dfrac{2x+3\sqrt{x}+9}{x-9}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(\Rightarrow B=\dfrac{2x+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{2x+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{x+6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Rightarrow B=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

Answer 2

\(\dfrac{2x+3\sqrt{x}+9-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

Answer 3

\(B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)