Đáp án:
b. \(\dfrac{{2\sqrt 2 + 2}}{{3\sqrt 2 + 4}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:a \ge 0;a \ne 1\\
Q = \left( {\dfrac{{\sqrt a }}{{\sqrt a + 1}} - \dfrac{1}{{a - \sqrt a }}} \right):\left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{2}{{a - 1}}} \right)\\
= \left[ {\dfrac{{a\left( {\sqrt a - 1} \right) - \sqrt a - 1}}{{\sqrt a \left( {a - 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt a - 1 + 2}}{{a - 1}}} \right]\\
= \dfrac{{a\sqrt a - a - \sqrt a - 1}}{{\sqrt a \left( {a - 1} \right)}}.\dfrac{{a - 1}}{{\sqrt a + 1}}\\
= \dfrac{{a\sqrt a - a - \sqrt a - 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}} = \dfrac{{a\sqrt a - a - \sqrt a - 1}}{{a + \sqrt a }}\\
b.Thay:a = 3 + 2\sqrt 2 = 2 + 2\sqrt 2 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to Q = \dfrac{{\left( {3 + 2\sqrt 2 } \right)\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 3 - 2\sqrt 2 - \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 1}}{{3 + 2\sqrt 2 + \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}\\
= \dfrac{{\left( {3 + 2\sqrt 2 } \right)\left( {\sqrt 2 + 1} \right) - 3 - 2\sqrt 2 - \sqrt 2 - 1 - 1}}{{3 + 2\sqrt 2 + \sqrt 2 + 1}}\\
= \dfrac{{3\sqrt 2 + 3 + 4 + 2\sqrt 2 - 3\sqrt 2 - 5}}{{3\sqrt 2 + 4}}\\
= \dfrac{{2\sqrt 2 + 2}}{{3\sqrt 2 + 4}}\\
c.Q < 0\\
\to \dfrac{{a\sqrt a - a - \sqrt a - 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}} < 0\\
\to a\sqrt a - a - \sqrt a - 1 < 0\\
\left( {Do:\sqrt a \left( {\sqrt a + 1} \right) > 0\forall a \ge 0} \right)\\
\to \sqrt a < 1,839286755\\
\to 0 \le a < 3,382975767;a \ne 1
\end{array}\)
