ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x-1}-\sqrt{2}\ne0\\x\ge1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ge1\end{matrix}\right.\)
a) \(D=\frac{x-3}{\left(\frac{x-1-2}{\sqrt{x-1}+\sqrt{2}}\right)}=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-3}=\sqrt{x-1}+\sqrt{2}\)
b) Thay x = \(4\left(2-\sqrt{3}\right)\) vào biểu thức ta được:
\(D=\sqrt{x-1}+\sqrt{2}=\sqrt{7-4\sqrt{3}}+\sqrt{2}=\sqrt{4-2.2.\sqrt{3}+3}+\sqrt{2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{2}\) . Nhận xét rằng \(2-\sqrt{3}=\sqrt{4}-\sqrt{3}>0\). Do đó:
\(D=2-\sqrt{3}+\sqrt{2}\)
c) \(D=\sqrt{x-1}+\sqrt{2}\ge0+\sqrt{2}=\sqrt{2}\)
Đẳng thức xảy ra khi x = 1
