Đáp án:
b. a>1
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a \ge 0;a \ne \pm 1\\
A = \frac{{1 - \sqrt a + 1 + \sqrt a }}{{2\left( {1 - a} \right)}} - \frac{{{a^2} + 1}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}\\
= \frac{{2 + 2a - 2{a^2} - 2}}{{2\left( {1 - a} \right)\left( {1 + a} \right)}}\\
= \frac{{2a\left( {1 - a} \right)}}{{2\left( {1 - a} \right)\left( {1 + a} \right)}}\\
= \frac{a}{{1 + a}}\\
b.A < \frac{1}{3}\\
\to \frac{a}{{1 + a}} < \frac{1}{3}\\
\to \frac{{3a - 1 - a}}{{1 - a}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2a - 1 > 0\\
1 - a > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2a - 1 < 0\\
1 - a < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
a > 1\\
\left\{ \begin{array}{l}
a < \frac{1}{2}\\
a > 1
\end{array} \right.\left( l \right)
\end{array} \right.
\end{array}\)
