a: ĐKXĐ: a>=0
\(a-\sqrt{a}-2=2\)
=>\(a-\sqrt{a}-4=0\)
=>\(a-\sqrt{a}+\frac14-\frac{17}{4}=0\)
=>\(\left(\sqrt{a}-\frac12\right)^2=\frac{17}{4}\)
=>\(\left[\begin{array}{l}\sqrt{a}-\frac12=\frac{\sqrt{17}}{2}\\ \sqrt{a}-\frac12=-\frac{\sqrt{17}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{a}=\frac{\sqrt{17}+1}{2}\\ \sqrt{a}=\frac{-\sqrt{17}+1}{2}\left(loại\right)\end{array}\right.\)
=>\(\sqrt{a}=\frac{\sqrt{17}+1}{2}\)
=>\(a=\frac{\left(\sqrt{17}+1\right)^2}{4}=\frac{18+2\sqrt{17}}{4}=\frac{9+\sqrt{17}}{2}\) (nhận)
b: \(E=a-\sqrt{a}-2\)
\(=a-\sqrt{a}+\frac14-\frac94\)
\(=\left(\sqrt{a}-\frac12\right)^2-\frac94\ge-\frac94\forall a\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{a}-\frac12=0\)
=>\(\sqrt{a}=\frac12\)
=>\(a=\frac14\) (nhận)
