\(a,\) ĐKXĐ\(:\) \(a\) ≥ \(0\) \(;\) \(a\) ≠ \(4\) \(;\) \(a\) ≠ \(9\)
\(M=\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}+\dfrac{\sqrt{a}+2}{3-\sqrt{a}}+\dfrac{\sqrt{a}+2}{a-5\sqrt{a}+6}\right)\\ =\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}+\dfrac{\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\right)\\ =\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{a-9-a+4+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\right)\\ =\dfrac{1}{\sqrt{a}+1}:\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\\ =\dfrac{1}{\sqrt{a}+1}:\dfrac{1}{\sqrt{a}-2}\\ =\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\)
\(b,\) Với \(a\) ≥ \(0\) \(;\) \(a\) ≠ \(4\) \(;\) \(a\) ≠ \(9\) ta có\(:\)
\(M< 0\) ⇔ \(\dfrac{\sqrt{a}-2}{\sqrt{a}+1}< 0\)
Mà\(:\) \(\sqrt{a}+1>0\) với ∀ \(a\) ≥ \(0\) \(;\) \(a\) ≠ \(4\) \(;\) \(a\) ≠ \(9\)
⇒\(M< 0\) ⇔ \(\sqrt{a}-2< 0\)
⇔ \(\sqrt{a}< 2\)
⇔ \(a< 4\)
Kết hợp với ĐKXĐ ta có\(:\) \(0\) ≤ \(a< 4\)
Vậy \(0\) ≤ \(a< 4\) thì \(M< 0\)
