Biểu thức $P = \left( {\dfrac{{a – \sqrt a }}{{\sqrt a – 1}} – \dfrac{{\sqrt a + 1}}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a + 1}}{a}$ có nghĩa khi và chỉ khi $\left\{ \begin{array}{l} a \ge 0\\ a \ne 1\\ a \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a > 0\\ a \ne 1 \end{array} \right.$
Vậy $P$ có nghĩa $ \Leftrightarrow a > 0;a \ne 1$
b) ĐK: $a > 0;a \ne 1$
Ta có:
$\begin{array}{l} P = \left( {\dfrac{{a – \sqrt a }}{{\sqrt a – 1}} – \dfrac{{\sqrt a + 1}}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a + 1}}{a}\\ = \left( {\dfrac{{\sqrt a \left( {\sqrt a – 1} \right)}}{{\sqrt a – 1}} – \dfrac{{\sqrt a + 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right).\dfrac{a}{{\sqrt a + 1}}\\ = \left( {\sqrt a – \dfrac{1}{{\sqrt a }}} \right).\dfrac{a}{{\sqrt a + 1}}\\ = \dfrac{{a – 1}}{{\sqrt a }}.\dfrac{a}{{\sqrt a + 1}}\\ = \dfrac{{\left( {\sqrt a – 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a }}.\dfrac{a}{{\sqrt a + 1}}\\ = \sqrt a \left( {\sqrt a – 1} \right) \end{array}$
Vậy $P = \sqrt a \left( {\sqrt a – 1} \right)$ với $a > 0;a \ne 1$
c) Ta có: Từ câu b với $a > 0;a \ne 1$
$\begin{array}{l} P = \sqrt a \left( {\sqrt a – 1} \right)\\ = {\left( {\sqrt a } \right)^2} – \sqrt a \\ = {\left( {\sqrt a – \dfrac{1}{2}} \right)^2} – \dfrac{1}{4} \end{array}$
Mà ${\left( {\sqrt a – \dfrac{1}{2}} \right)^2} \ge 0,\dforall a > 0;a \ne 1$
$\begin{array}{l} \Rightarrow P = {\left( {\sqrt a – \dfrac{1}{2}} \right)^2} – \dfrac{1}{4} \ge \dfrac{{ – 1}}{4},\dforall a > 0;a \ne 1\\ \Rightarrow MinP = \dfrac{{ – 1}}{4} \end{array}$
Dấu bằng xảy ra
$\begin{array}{l} {\left( {\sqrt a – \dfrac{1}{2}} \right)^2} = 0\\ \Leftrightarrow \sqrt a – \dfrac{1}{2} = 0\\ \Leftrightarrow \sqrt a = \dfrac{1}{2}\\ \Leftrightarrow a = \dfrac{1}{4} \end{array}$
Vậy $MinP = \dfrac{{ – 1}}{4} \Leftrightarrow a = \dfrac{1}{4}$
Giải thích các bước giải:
a) Ta có:
Biểu thức $P = \left( {\dfrac{{a – \sqrt a }}{{\sqrt a – 1}} – \dfrac{{\sqrt a + 1}}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a + 1}}{a}$ có nghĩa khi và chỉ khi $\left\{ \begin{array}{l}
a \ge 0\\
a \ne 1\\
a \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
a \ne 1
\end{array} \right.$
Vậy $P$ có nghĩa $ \Leftrightarrow a > 0;a \ne 1$
b) ĐK: $a > 0;a \ne 1$
Ta có:
$\begin{array}{l}
P = \left( {\dfrac{{a – \sqrt a }}{{\sqrt a – 1}} – \dfrac{{\sqrt a + 1}}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a + 1}}{a}\\
= \left( {\dfrac{{\sqrt a \left( {\sqrt a – 1} \right)}}{{\sqrt a – 1}} – \dfrac{{\sqrt a + 1}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right).\dfrac{a}{{\sqrt a + 1}}\\
= \left( {\sqrt a – \dfrac{1}{{\sqrt a }}} \right).\dfrac{a}{{\sqrt a + 1}}\\
= \dfrac{{a – 1}}{{\sqrt a }}.\dfrac{a}{{\sqrt a + 1}}\\
= \dfrac{{\left( {\sqrt a – 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a }}.\dfrac{a}{{\sqrt a + 1}}\\
= \sqrt a \left( {\sqrt a – 1} \right)
\end{array}$
Vậy $P = \sqrt a \left( {\sqrt a – 1} \right)$ với $a > 0;a \ne 1$
c) Ta có: Từ câu b với $a > 0;a \ne 1$
$\begin{array}{l}
P = \sqrt a \left( {\sqrt a – 1} \right)\\
= {\left( {\sqrt a } \right)^2} – \sqrt a \\
= {\left( {\sqrt a – \dfrac{1}{2}} \right)^2} – \dfrac{1}{4}
\end{array}$
Mà ${\left( {\sqrt a – \dfrac{1}{2}} \right)^2} \ge 0,\dforall a > 0;a \ne 1$
$\begin{array}{l}
\Rightarrow P = {\left( {\sqrt a – \dfrac{1}{2}} \right)^2} – \dfrac{1}{4} \ge \dfrac{{ – 1}}{4},\dforall a > 0;a \ne 1\\
\Rightarrow MinP = \dfrac{{ – 1}}{4}
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
{\left( {\sqrt a – \dfrac{1}{2}} \right)^2} = 0\\
\Leftrightarrow \sqrt a – \dfrac{1}{2} = 0\\
\Leftrightarrow \sqrt a = \dfrac{1}{2}\\
\Leftrightarrow a = \dfrac{1}{4}
\end{array}$
Vậy $MinP = \dfrac{{ – 1}}{4} \Leftrightarrow a = \dfrac{1}{4}$