Đáp án:
a) \(\dfrac{1}{{ - a + \sqrt a }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a \ge 0;a \ne 1\\
A = \dfrac{{1 + \sqrt a + 1 - \sqrt a }}{{1 - a}}:\dfrac{{1 + \sqrt a - 1 + \sqrt a }}{{1 - a}} + \dfrac{1}{{1 - \sqrt a }}\\
= \dfrac{2}{{1 - a}}.\dfrac{{1 - a}}{{2\sqrt a }} + \dfrac{1}{{1 - \sqrt a }}\\
= \dfrac{1}{{\sqrt a }} + \dfrac{1}{{1 - \sqrt a }} = \dfrac{{1 - \sqrt a + \sqrt a }}{{\sqrt a \left( {1 - \sqrt a } \right)}}\\
= \dfrac{1}{{ - a + \sqrt a }}\\
b)Thay:a = 7 + 4\sqrt 3 = 4 + 2.2.\sqrt 3 + 3\\
= {\left( {2 + \sqrt 3 } \right)^2}\\
\to A = \dfrac{1}{{ - 7 - 4\sqrt 3 + \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} }}\\
= \dfrac{1}{{ - 7 - 4\sqrt 3 + 2 + \sqrt 3 }} = \dfrac{1}{{ - 5 - 3\sqrt 3 }}
\end{array}\)
