a) Để A có nghĩa thì \(x-5\ge0\Rightarrow x\ge5\)
b) * \(A=0\)
\(\sqrt{x-5}\cdot A=\left(\sqrt{x-5}\right)^2\)
\(0\cdot\sqrt{x-5}=\left|x-5\right|\)
\(0=x-5\)
\(x=5\)
* \(A=4\)
\(A=\sqrt{x-5}\)
\(4=\sqrt{x-5}\)
\(\sqrt{16}=\sqrt{x-5}\)
Suy ra: \(16=x-5\)
\(x=21\)