Đáp án:
\(\begin{array}{l}
1,\\
B = \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
2,\\
A = \dfrac{1}{{\sqrt x - 3}}\\
3,\\
P = \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
4,\\
C = \dfrac{1}{{\sqrt a + 1}}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
DKXD:\,\,\,x > 0\\
B = \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x + 1}}{{x + \sqrt x }}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x + 1}}{{{{\sqrt x }^2} + \sqrt x }}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{{2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\sqrt x }^2} - {1^2} + 2\sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\sqrt x }^2} + 2\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
2,\\
A = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{3}{{\sqrt x - 3}}} \right).\dfrac{{\sqrt x + 3}}{{x + 9}}\\
= \dfrac{{\sqrt x .\left( {\sqrt x - 3} \right) + 3.\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{x + 9}}\\
= \dfrac{{x - 3\sqrt x + 3\sqrt x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{x + 9}}\\
= \dfrac{{x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x + 3}}{{x + 9}}\\
= \dfrac{1}{{\sqrt x - 3}}\\
3,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\\
P = \dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{1}{{x - \sqrt x }}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\sqrt x }^2} - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
4,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right.\\
C = \dfrac{{a - \sqrt a }}{{a - 1}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
= \dfrac{{{{\sqrt a }^2} - \sqrt a }}{{{{\sqrt a }^2} - {1^2}}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a .\left( {\sqrt a - 1} \right)}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a }}{{\sqrt a + 1}} - \dfrac{{\sqrt a - 1}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - \left( {\sqrt a - 1} \right)}}{{\sqrt a + 1}}\\
= \dfrac{1}{{\sqrt a + 1}}
\end{array}\)