a) ĐKXĐ: x ≥ 2
b) Ta có:
\(P=\dfrac{x-5}{\sqrt{x-2}-\sqrt{3}}\\ =\dfrac{\left(x-5\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{x-2-3}\\ =\sqrt{x-2}+\sqrt{3}\)
Vậy P = \(\sqrt{x-2}+\sqrt{3}\) với x ≥ 2
c) Để P = \(\sqrt{27}\)
\(\\ \Leftrightarrow\sqrt{x-2}+\sqrt{3}=\sqrt{27}=3\sqrt{3}\\ \Leftrightarrow\sqrt{x-2}=2\sqrt{2}\\ \Leftrightarrow x-2=8\\ \Leftrightarrow x=10\left(TM\right)\)
Vậy x =10
d) Ta có: \(P=\sqrt{x-2}+\sqrt{3}\)
Vì x ≥ 2 nên \(\sqrt{x-2}\ge0\Leftrightarrow\sqrt{x-2}+\sqrt{3}\ge\sqrt{3}\)
hay P ≥ √3
Dấu '=' xảy ra ⇔ x = 2 (TM)
Vậy MinP = √3 ⇔ x = 2