Giúp mình với:
Cho b/t A=\(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{x-9}\) (\(x\ge0\) ; \(x\ne9\) )
Rút gọn b/t A rồi tính giá trị tại x= \(2\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
Mai Kt rồi :( :( :(
a. \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\sqrt{x}+3}\)
. \(x=2.\left(4+\sqrt{15}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4-\sqrt{15}}\)
\(\Rightarrow x=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^3\)\(=4\left(\sqrt{5}-\sqrt{3}\right)\)
Thay \(x=4\left(\sqrt{5}-\sqrt{3}\right)\Rightarrow A=\frac{3}{\sqrt{4\left(\sqrt{5}-\sqrt{3}\right)}+3}\)
\(=\frac{3}{2\sqrt{\left(\sqrt{5}-\sqrt{3}\right)}+3}\)