a,Đk: a≥0 ; a khác 4
H=\(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\) -\(\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\) -\(\dfrac{1}{\sqrt{a}-2}\)
= \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
=\(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
=\(\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
=\(\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
=\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b, Để H<2
<=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) <2
<=> \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) -2<0
<=>\(\dfrac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}\) <0
<=>\(\dfrac{-\sqrt{a}}{\sqrt{a}-2}\) <0
<=>\(\left\{{}\begin{matrix}-\sqrt{a}< 0\\\sqrt{a}-2>0\end{matrix}\right.\) ( vì \(\sqrt{a}>0< =>-\sqrt{a}< 0\)
<=> a>4
vậy để H <2 khi a>4
c, Ta có a\(^2\) +3a=0
<=> a(a+3)=0
<=>a=0 hoặc a=-3(vô lí)
+ Với a=0 <=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) =\(\dfrac{0-4}{0-2}\) =2
d, Để H=5
<=> \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) =5
<=>\(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) -5=0
<=>\(\dfrac{\sqrt{a}-4-5\sqrt{a}+10}{\sqrt{a}-2}\) =0
<=>-4\(\sqrt{a}\) +6=0
<=> a=\(\dfrac{9}{4}\)
