Hãy rút gọn biểu thức: \(M={\left(\dfrac{1}{a -\sqrt a} +\dfrac{1}{\sqrt a -1}\right)} : \dfrac{\sqrt a +1}{a -2\sqrt a+1}\) với \(a > 0\

* Hướng dẫn giải

Ta có: 

\(M={\left(\dfrac{1}{a -\sqrt a} +\dfrac{1}{\sqrt a -1}\right)} : \dfrac{\sqrt a +1}{a -2\sqrt a+1}\)

      \(={\left(\dfrac{1}{\sqrt a .\sqrt a -\sqrt a .1}+\dfrac{1}{\sqrt a -1} \right)} : \dfrac{\sqrt a +1}{(\sqrt a)^2 -2\sqrt a+1}\)

      \(={\left(\dfrac{1}{\sqrt a(\sqrt a -1)}+\dfrac{1}{\sqrt a -1} \right)} : \dfrac{\sqrt a +1}{(\sqrt a -1)^2}\)

      \(={\left(\dfrac{1}{\sqrt a(\sqrt a -1)}+\dfrac{\sqrt a}{\sqrt a(\sqrt a -1)} \right)} : \dfrac{\sqrt a +1}{(\sqrt a -1)^2}\)

      \(=\dfrac{1+\sqrt a}{\sqrt a(\sqrt a -1)}  : \dfrac{\sqrt a +1}{(\sqrt a -1)^2}\)

     \(=\dfrac{1+\sqrt a}{\sqrt a(\sqrt a -1)}  . \dfrac{(\sqrt a -1)^2}{\sqrt a +1}\)

     \(=\dfrac{1}{\sqrt a}  . \dfrac{\sqrt a -1}{1}=\dfrac{\sqrt a -1}{\sqrt a}\).

     \(=\dfrac{\sqrt a}{\sqrt a}-\dfrac{1}{\sqrt a} =1 -\dfrac{1}{\sqrt a}\)