Giải thích các bước giải:
\(\begin{array}{l}
10,\\
a,\\
{\rm{DK:}}\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
2 + 2\sqrt a \ne 0\\
2 - 2\sqrt a \ne 0\\
1 - {a^2} \ne 0\\
a \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.\\
b,\\
Q = \left( {\dfrac{1}{{2 + 2\sqrt a }} + \dfrac{1}{{2 - 2\sqrt a }} - \dfrac{{{a^2} + 1}}{{1 - {a^2}}}} \right).\left( {1 + \dfrac{1}{a}} \right)\\
= \left( {\dfrac{1}{{2\left( {1 + \sqrt a } \right)}} + \dfrac{1}{{2.\left( {1 - \sqrt a } \right)}} - \dfrac{{{a^2} + 1}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}} \right).\dfrac{{a + 1}}{a}\\
= \left( {\dfrac{{\left( {1 - \sqrt a } \right) + \left( {1 + \sqrt a } \right)}}{{2\left( {1 + \sqrt a } \right).\left( {1 - \sqrt a } \right)}} - \dfrac{{{a^2} + 1}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}} \right).\dfrac{{a + 1}}{a}\\
= \left( {\dfrac{2}{{2.\left( {1 + \sqrt a } \right).\left( {1 - \sqrt a } \right)}} - \dfrac{{{a^2} + 1}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}} \right).\dfrac{{a + 1}}{a}\\
= \left( {\dfrac{1}{{1 - a}} - \dfrac{{{a^2} + 1}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}} \right).\dfrac{{a + 1}}{a}\\
= \dfrac{{a + 1 - \left( {{a^2} + 1} \right)}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}.\dfrac{{a + 1}}{a}\\
= \dfrac{{a - {a^2}}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}.\dfrac{{a + 1}}{a}\\
= \dfrac{{a.\left( {1 - a} \right)}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}.\dfrac{{a + 1}}{a}\\
= 1\\
11,\\
DK:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
y > 0\\
x \ne 4y\\
x \ne 1
\end{array} \right.\\
a,\\
A = \dfrac{{\sqrt {{x^3}} }}{{\sqrt {xy} - 2y}} + \dfrac{{2x}}{{2\sqrt {xy} + 2\sqrt y - x - \sqrt x }}.\dfrac{{1 - x}}{{1 - \sqrt x }}\\
= \dfrac{{x\sqrt x }}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}} + \dfrac{{2x}}{{2\sqrt y \left( {\sqrt x + 1} \right) - \sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{1 - \sqrt x }}\\
= \dfrac{{x\sqrt x }}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}} + \dfrac{{2x}}{{\left( {\sqrt x + 1} \right).\left( {2\sqrt y - \sqrt x } \right)}}.\left( {1 + \sqrt x } \right)\\
= \dfrac{{x\sqrt x }}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}} + \dfrac{{2x}}{{2\sqrt y - \sqrt x }}\\
= \dfrac{{x\sqrt x - 2x\sqrt y }}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}}\\
= \dfrac{{x\left( {\sqrt x - 2\sqrt y } \right)}}{{\sqrt y \left( {\sqrt x - 2\sqrt y } \right)}}\\
= \dfrac{x}{{\sqrt y }}\\
b,\\
y = 625 \Rightarrow \sqrt y = 25\\
A < 0,2 \Leftrightarrow \dfrac{x}{{25}} < 0,2 \Leftrightarrow x < 5
\end{array}\)
