Đáp án:
\(\begin{array}{l}
C3:\\
a)A = \sqrt a - 1\\
C4:\\
A = \dfrac{1}{{\sqrt x - 2}}\\
B = 1\\
C5:\\
P = \dfrac{{2\left( {a - 2} \right)}}{{a + 2}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C3:\\
a)A = \left[ {\dfrac{{\sqrt a }}{{\sqrt a - 1}} - \dfrac{{\sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right]:\dfrac{{\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \left( {\dfrac{{\sqrt a }}{{\sqrt a - 1}} - \dfrac{1}{{\sqrt a - 1}}} \right).\dfrac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a - 1}}.\left( {\sqrt a - 1} \right) = \sqrt a - 1\\
C4:\\
A = \left[ {\dfrac{{3\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} + \dfrac{{\sqrt x }}{{\sqrt x - 2}}} \right].\dfrac{{\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3 + \sqrt x }}{{\sqrt x - 2}}.\dfrac{{\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{1}{{\sqrt x - 2}}\\
B = \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right].{\left[ {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right]^2}\\
= \left( {1 + \sqrt a + a + \sqrt a } \right).\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}}\\
= {\left( {1 + \sqrt a } \right)^2}.\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}} = 1\\
C5:\\
P = \left[ {\dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right].\dfrac{{a - 2}}{{a + 2}}\\
= \left( {\dfrac{{a + \sqrt a + 1 - a + \sqrt a - 1}}{{\sqrt a }}} \right).\dfrac{{a - 2}}{{a + 2}}\\
= \dfrac{{2\sqrt a }}{{\sqrt a }}.\dfrac{{a - 2}}{{a + 2}} = \dfrac{{2\left( {a - 2} \right)}}{{a + 2}}
\end{array}\)
