Đáp án:
a) \(\dfrac{{\sqrt x - 2}}{{3\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{{\sqrt x - \sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right]:\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}} \right]\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}{{x - 1 - x + 4}}\\
= \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
b)A = - \dfrac{2}{3}\\
\to \dfrac{{\sqrt x - 2}}{{3\sqrt x }} = - \dfrac{2}{3}\\
\to 3\sqrt x - 6 = - 6\sqrt x \\
\to 9\sqrt x = 6\\
\to \sqrt x = \dfrac{2}{3}\\
\to x = \dfrac{4}{9}\\
c)A = \dfrac{{\sqrt x - 2}}{{3\sqrt x }}\\
\to 3A = \dfrac{{\sqrt x - 2}}{{\sqrt x }} = 1 - \dfrac{2}{{\sqrt x }}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x }} \in Z\\
\Leftrightarrow \sqrt x \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 1
\end{array} \right. \to \left[ \begin{array}{l}
A = 0\left( l \right)\\
A = - 1\left( l \right)
\end{array} \right.
\end{array}\)
⇒ Không có giá trị x để A là số nguyên dương
