a, \(\left(\frac{1}{x+2\sqrt{x}}-\frac{1}{\sqrt{x}+2}\right):\frac{1-\sqrt{x}}{x+4\sqrt{x}+4}\)ĐK : x >= 0 ; \(x\ne1\)
\(=\left(\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
b, \(F=\frac{5}{2}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\)
ĐK : x > 0 , x khác 1
\(bthuc=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}}\)
Để bthuc = 5/2 thì \(\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{5}{2}\Rightarrow2\sqrt{x}+4=5\sqrt{x}\Leftrightarrow3\sqrt{x}=4\Leftrightarrow x=\frac{16}{9}\left(tm\right)\)
Điều kiện Quỳnh mới đúng nhé, quên cái căn x kia :v