Đáp án:
\(\begin{array}{l}
a,\\
Q = \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
b,\\
x > 4
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
Q = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{{\sqrt x }}} \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\\
= \dfrac{{\sqrt x - \left( {\sqrt x - 1} \right)}}{{\sqrt x .\left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - \sqrt x + 1}}{{\sqrt x .\left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {{{\sqrt x }^2} - {1^2}} \right) - \left( {{{\sqrt x }^2} - {2^2}} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {x - 1} \right) - \left( {x - 4} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{3}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{1}{{\sqrt x .\left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{3}\\
= \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
b,\\
Q > 0\\
\Leftrightarrow \dfrac{{\sqrt x - 2}}{{3\sqrt x }} > 0\\
3\sqrt x > 0,\,\,\,\forall x > 0,x \ne 4,x \ne 1\\
\Rightarrow \sqrt x - 2 > 0\\
\Leftrightarrow \sqrt x > 2\\
\Leftrightarrow x > {2^2}\\
\Leftrightarrow x > 4
\end{array}\)
