Đáp án:
\(\begin{array}{l}
A = \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
B = 2\sqrt a \\
C = \dfrac{{4a - 1}}{{{a^2}}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne 1\\
A = \dfrac{{1 + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
DK:a > 0;a \ne 1\\
B = 2\left( {\dfrac{{\sqrt a - \sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right).\dfrac{{a\left( {a - 1} \right)}}{{\sqrt a + 1}}\\
= \dfrac{2}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{a\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{\sqrt a + 1}}\\
= 2\sqrt a \\
DK:a > 0;a \ne 1\\
\dfrac{{4a\sqrt a - \sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a - 1}}{{{a^2}}}\\
= \dfrac{{4a - 1}}{{\sqrt a - 1}}.\dfrac{{\sqrt a - 1}}{{{a^2}}}\\
= \dfrac{{4a - 1}}{{{a^2}}}
\end{array}\)
