Đáp án:
b) a>4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a > 0;a \ne \left\{ {1;4} \right\}\\
Q = \left[ {\dfrac{{\sqrt a - \sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right]:\left[ {\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right) - \left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)}}{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}} \right]\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a - 1} \right)}}{{a - 1 - a + 4}}\\
= \dfrac{{\sqrt a - 2}}{{3\sqrt a }}\\
b)Q > 0\\
\to \dfrac{{\sqrt a - 2}}{{3\sqrt a }} > 0\\
\to \sqrt a - 2 > 0\left( {do:\sqrt a > 0\forall a > 0} \right)\\
\to a > 4\\
c)Thay:a = 9 - 4\sqrt 5 \\
= 5 - 2.2\sqrt 5 + 4 = {\left( {\sqrt 5 - 2} \right)^2}\\
\to Q = \dfrac{{\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - 2}}{{3\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} }} = \dfrac{{\sqrt 5 - 2 - 2}}{{3\left( {\sqrt 5 - 2} \right)}}\\
= \dfrac{{\sqrt 5 - 4}}{{3\sqrt 5 - 6}} = \dfrac{{ - 3 - 2\sqrt 5 }}{3}
\end{array}\)
