a) ĐKXĐ :
\(\hept{\begin{cases}a\ge0\\a\ne4\end{cases}}\)
b) Với \(a\ge0\) và \(a\ne4\)
\(A=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\frac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)
Để A > 2
thì \(\frac{\sqrt{a}-4}{\sqrt{a}-2}>2\)
Ta có :
\(\frac{\sqrt{a}-4}{\sqrt{a}-2}-2\)
\(=\frac{\sqrt{a}-4-2\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=\frac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}\)
\(\)\(=\frac{-\sqrt{a}}{\sqrt{a}-2}\)
+) \(-\sqrt{a}< 0\forall a\) \(\Rightarrow a>0\)
+) \(\sqrt{a}-2< 0\) \(\Leftrightarrow a< 4\)
Vậy để A > 2 thì 0 < a < 4
c) Để A = 5
thì \(\frac{\sqrt{a}-4}{\sqrt{a}-2}=5\)
\(\frac{\left(\sqrt{a}-4\right)-5\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)}=0\)
\(\frac{\sqrt{a}-4-5\sqrt{a}+10}{\sqrt{a}-2}=0\)
\(\Rightarrow-4\sqrt{a}+6=0\)
\(\Rightarrow a=\frac{9}{4}\)( TMĐKXĐ )
Vậy để A = 5 thì a = 9/4
a, A xđ <=> \(\hept{\begin{cases}\sqrt{a}+3\ne0\\a+\sqrt{a}-6\ne0\\2-\sqrt{a}\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne2\\a\ne4\end{cases}};a\ne-3\)-3
b, rút gọn: A=\(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)để A> 2 <=> \(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)>2 <=> 1+\(\frac{-2}{\sqrt{a}-2}\)>2 <=> \(\frac{\sqrt{a}}{2-\sqrt{a}}\)>0
mà a\(\ge\)0 <=> \(\sqrt{a}\ge0\)=> \(2-\sqrt{a}\)>0 <=> a<4
kết hợp với điều kiện, ta được: \(0\le a< 4;a\ne2\)
c, để A = 5 thì \(\frac{-2}{\sqrt{a}-2}\)+1=5
<=> \(\frac{-2}{\sqrt{a}-2}\)=4
<=> \(a=\frac{9}{4}\)(t/m)
KL..............
