Ta có: \(\left(\frac{4-2\sqrt{a}}{2-\sqrt{a}}-\frac{a-\sqrt{a}}{1-\sqrt{a}}\right)\cdot\frac{4-a}{2-\sqrt{a}}+a\)
\(=\left(\frac{2\left(2-\sqrt{a}\right)}{2-\sqrt{a}}+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\cdot\frac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}+a\)
\(=\left(2+\sqrt{a}\right)\left(2+\sqrt{a}\right)+a\)
\(=a+4\sqrt{a}+4+a\)
\(=2a+4\sqrt{a}+4\)
= \(\left[\frac{2\left(2-\sqrt{a}\right)}{2-\sqrt{a}}-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{1-\sqrt{a}}\right]\). \(\left[\frac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}+a\right]\)
= \(\left[2+\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right]\). \(\left(2+\sqrt{a}\right)+a\)
= \(\left(2+\sqrt{a}\right).\left(2+\sqrt{a}\right)+a\)
= \(\left(2+\sqrt{a}\right)^2+a\)
= \(4+4\sqrt{a}+a+a\)
= \(4+4\sqrt{a}+2a\)
