Question

Cho A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+9}{x-9}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) (x>0, x≠9)

a)Rút gọn A

b) Tìm x sao cho A<-1

Answer 1

Answer 2

a, \(\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):(\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}+3\right)})\)

\(=\frac{-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{2\sqrt{x}+4}\)

\(=\frac{-3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

câu b mình quên cách tính rồi sorry