Đáp án:
d) a>4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a > 0;a \ne \left\{ {1;4} \right\}\\
b)M = \left[ {\dfrac{{\sqrt a - \sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right]:\left[ {\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a - 2} \right)}}} \right]\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a - 2} \right)}}{{a - 1 - a + 4}}\\
= \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\\
c)M = \dfrac{1}{2}\\
\to \dfrac{{\sqrt a - 2}}{{3\sqrt a }} = \dfrac{1}{2}\\
\to 2\sqrt a - 4 = 3\sqrt a \\
\to \sqrt x = - 4\left( l \right)\\
\to x \in \emptyset \\
d)M > 0\\
\to \dfrac{{\sqrt a - 2}}{{3\sqrt a }} > 0\\
\to \sqrt a - 2 > 0\left( {do:3\sqrt a > 0\forall a > 0} \right)\\
\to a > 4
\end{array}\)
