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$(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a})(\frac{1}{1+2\sqrt{a}+a})=(\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a})·\frac{1}{1+2\sqrt{a}+a}$

$=(1+\sqrt{a}+a+\sqrt{a})·\frac{1}{1+2\sqrt{a}+a}=(1+2\sqrt{a}+a)\frac{1}{1+2\sqrt{a}+a}=1$