Đáp án:
B6: \(\sqrt a + 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
P = \left[ {\dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right].\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{1 - x}}{x}\\
B3:\\
P = \left[ {\dfrac{{\sqrt a + 3 + \sqrt a - 3}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)}}} \right].\left( {\dfrac{{\sqrt a - 3}}{{\sqrt a }}} \right)\\
= \dfrac{{2\sqrt a }}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 3} \right)}}\left( {\dfrac{{\sqrt a - 3}}{{\sqrt a }}} \right)\\
= \dfrac{2}{{\sqrt a + 3}}\\
B4:\\
P = \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\\
= \dfrac{{1 - x}}{x}\\
B5:\\
A = \left[ {\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x - 1 + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{x - 1}}{{\sqrt x }}\\
B6:\\
A = \dfrac{{1 + a - 2\sqrt a }}{{a + 1}}:\left[ {\dfrac{1}{{\sqrt a + 1}} - \dfrac{{2\sqrt a }}{{a\left( {\sqrt a + 1} \right) + \left( {\sqrt a + 1} \right)}}} \right]\\
= \dfrac{{1 + a - 2\sqrt a }}{{a + 1}}:\left[ {\dfrac{1}{{\sqrt a + 1}} - \dfrac{{2\sqrt a }}{{\left( {\sqrt a + 1} \right)\left( {a + 1} \right)}}} \right]\\
= \dfrac{{1 + a - 2\sqrt a }}{{a + 1}}:\left[ {\dfrac{{a + 1 - 2\sqrt a }}{{\left( {\sqrt a + 1} \right)\left( {a + 1} \right)}}} \right]\\
= \dfrac{{1 + a - 2\sqrt a }}{{a + 1}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {a + 1} \right)}}{{a + 1 - 2\sqrt a }}\\
= \sqrt a + 1
\end{array}\)
