от Nathi123 » 07 Мар 2016, 13:26
A(x) = [tex]\frac{1+\sqrt{x}}{x\sqrt{x}+2x+\sqrt{x}}:\frac{1}{\sqrt{x}} = \frac{1+\sqrt{x}}{\sqrt{x}(x+2\sqrt{x}+1)}.(1+\sqrt{x})[/tex] при x>0[tex]\Rightarrow A(x) = \frac{(1+\sqrt{x})^{2}}{\sqrt{x}(\sqrt{x}+1)^{2}}=\frac{1}{\sqrt{x}} \Rightarrow A(9)=\frac{1}{3}[/tex]