a: \(T=\left(1+\sqrt{a}+a+\sqrt{a}\right)\cdot\dfrac{1}{1+\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}=\sqrt{a}+1\)
b: Để T=6a thì \(6a-\sqrt{a}-1=0\)
\(\Leftrightarrow6a-3\sqrt{a}+2\sqrt{a}-1=0\)
\(\Leftrightarrow\left(2\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)=0\)
=>a=1/4