Đáp án:
\(\begin{array}{l}
a)\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1\\
x \ne 4
\end{array} \right..\\
b)\,\,\,P = \frac{{\sqrt x - 2}}{{3\sqrt x }}.\\
c)\,\,x > 4.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \left( {\frac{1}{{\sqrt x - 1}} - \frac{1}{{\sqrt x }}} \right):\frac{3}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
a)\,\,DK:\,\,\,\left\{ \begin{array}{l}
x > 0\\
\sqrt x - 1 \ne 0\\
\sqrt x - 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 1\\
x \ne 4
\end{array} \right..\\
b)\,\,\,P = \frac{{\sqrt x - \sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\frac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{3}\\
\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt x }}.\frac{{\sqrt x - 2}}{3} = \frac{{\sqrt x - 2}}{{3\sqrt x }}.\\
c)\,\,DK:\,\,\,x > 0,\,\,x \ne 1,\,\,x \ne 4.\\
P > 0 \Leftrightarrow \frac{{\sqrt x - 2}}{{3\sqrt x }} > 0\\
\Leftrightarrow \sqrt x - 2 > 0\,\,\,\left( {do\,\,\,3\sqrt x > 0\,\,\forall x\,\,tm\,\,dkxd} \right).\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > 4.\\
Vay\,\,x > 4\,\,\,thi\,\,P > 0.
\end{array}\)
