Đáp án:
10C 11B 12C
Giải thích các bước giải:
\(\begin{array}{l}
C10:\\
M = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x + 1}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
Xét:M - 1 = \dfrac{{\sqrt x - 1}}{{\sqrt x }} - 1\\
= \dfrac{{\sqrt x - 1 - \sqrt x }}{{\sqrt x }} = - \dfrac{1}{{\sqrt x }}\\
Do:x > 0 \to \sqrt x > 0\\
\to \dfrac{1}{{\sqrt x }} > 0\\
\to - \dfrac{1}{{\sqrt x }} < 0\\
\to M < 1\\
\to C\\
C11:\\
A = \dfrac{{\sqrt x + \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}:\dfrac{{2\sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{2\sqrt x + 3}}\\
= \dfrac{1}{{\sqrt x + 3}}\\
Xét:{A^2} - A = A\left( {A - 1} \right)\\
Do:\dfrac{1}{{\sqrt x + 3}} > 0\forall x \ge 0\\
\to Xet:A - 1 = \dfrac{1}{{\sqrt x + 3}} - 1\\
= \dfrac{{1 - \sqrt x - 3}}{{\sqrt x + 3}}\\
= \dfrac{{ - 2 - \sqrt x }}{{\sqrt x + 3}} = - \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}}\\
Do:\dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} > 0\forall x \ge 0\\
\to - \dfrac{{\sqrt x + 2}}{{\sqrt x + 3}} < 0\\
\to A - 1 < 0\\
\to {A^2} - A < 0\\
\to {A^2} < A\\
\to B\\
C12:\\
Q = \left[ {\dfrac{{\left( {1 - \sqrt x } \right)\left( {x + \sqrt x + 1} \right)}}{{1 - \sqrt x }} + \sqrt x } \right].\dfrac{{{{\left( {1 - \sqrt x } \right)}^2}}}{{{{\left[ {\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)} \right]}^2}}}\\
= \left( {x + \sqrt x + 1 + \sqrt x } \right).\dfrac{1}{{{{\left( {1 + \sqrt x } \right)}^2}}}\\
= {\left( {1 + \sqrt x } \right)^2}.\dfrac{1}{{{{\left( {1 + \sqrt x } \right)}^2}}}\\
= 1\\
\to C
\end{array}\)
