P=\(\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)(a ≥ 0; a≠4)
=\(\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+3\sqrt{a}-2\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)
=\(\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{1}{\sqrt{a}-2}\)
=\(\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\frac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\frac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\frac{a-4\sqrt{a}+3\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)
Ta có:
P<1
⇔\(\frac{\sqrt{a-4}}{\sqrt{a}-2}\)<1
⇔\(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)-1<0
⇔\(\frac{\sqrt{a}-4}{\sqrt{a}-2}-\frac{\sqrt{a}-2}{\sqrt{a}-2}\)<0
⇔\(\frac{-2}{\sqrt{a}-2}\) <0
Vì \(\frac{-2}{\sqrt{a}-2}\)<0 ⇒-2 và\(\sqrt{a}-2\) phải trái dấu. Mà -2 <0
⇒\(\sqrt{a}-2\) >0
⇔\(\sqrt{a}\) >2
⇔a >4
Vậy a> 4 để P>1
