Ta có:
\(M={\left(\dfrac{1}{a -\sqrt a} +\dfrac{1}{\sqrt a -1}\right)} : \dfrac{\sqrt a +1}{a -2\sqrt a+1}\)
\(={\left(\dfrac{1}{\sqrt a .\sqrt a -\sqrt a .1}+\dfrac{1}{\sqrt a -1} \right)} : \dfrac{\sqrt a +1}{(\sqrt a)^2 -2\sqrt a+1}\)
\(={\left(\dfrac{1}{\sqrt a(\sqrt a -1)}+\dfrac{1}{\sqrt a -1} \right)} : \dfrac{\sqrt a +1}{(\sqrt a -1)^2}\)
\(={\left(\dfrac{1}{\sqrt a(\sqrt a -1)}+\dfrac{\sqrt a}{\sqrt a(\sqrt a -1)} \right)} : \dfrac{\sqrt a +1}{(\sqrt a -1)^2}\)
\(=\dfrac{1+\sqrt a}{\sqrt a(\sqrt a -1)} : \dfrac{\sqrt a +1}{(\sqrt a -1)^2}\)
\(=\dfrac{1+\sqrt a}{\sqrt a(\sqrt a -1)} . \dfrac{(\sqrt a -1)^2}{\sqrt a +1}\)
\(=\dfrac{1}{\sqrt a} . \dfrac{\sqrt a -1}{1}=\dfrac{\sqrt a -1}{\sqrt a}\).
\(=\dfrac{\sqrt a}{\sqrt a}-\dfrac{1}{\sqrt a} =1 -\dfrac{1}{\sqrt a}\)
Vì \(a > 0 \Rightarrow \sqrt a > 0 \Rightarrow \dfrac{1}{\sqrt a} > 0 \Rightarrow 1 -\dfrac{1}{\sqrt a} < 1\).
Vậy \(M < 1\).
