Đáp án:
\(\begin{array}{l}
12,\\
A = \dfrac{x}{{\sqrt x + 1}}\\
13,\\
A = 1 - \sqrt x
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
12,\\
A = \left( {\dfrac{x}{{\sqrt x - 1}} - \sqrt x } \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} - \dfrac{1}{{1 - \sqrt x }} + \dfrac{{2 - x}}{{x - \sqrt x }}} \right)\\
= \dfrac{{x - \sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}:\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} + \dfrac{1}{{\sqrt x - 1}} + \dfrac{{2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right)\\
= \dfrac{{x - x + \sqrt x }}{{\sqrt x - 1}}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \sqrt x + \left( {2 - x} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}:\dfrac{{x - 1 + \sqrt x + 2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}:\dfrac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{{{\sqrt x }^2}}}{{\sqrt x + 1}}\\
= \dfrac{x}{{\sqrt x + 1}}\\
13,\\
A = \left( {\dfrac{{\sqrt x - 4}}{{x - 2\sqrt x }} - \dfrac{3}{{2 - \sqrt x }}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x - 2}}} \right)\\
= \left( {\dfrac{{\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}} + \dfrac{3}{{\sqrt x - 2}}} \right):\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - {{\sqrt x }^2}}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 4} \right) + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\dfrac{{\left( {x - 4} \right) - x}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\dfrac{{ - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4.\left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{ - 4}}\\
= \dfrac{{\sqrt x - 1}}{{ - 1}}\\
= 1 - \sqrt x
\end{array}\)
