Đáp án:
a) \(\dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}}\)
b) x=1
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 2}} - \dfrac{{x + 4}}{{x - 4}}} \right):\left( {\dfrac{{2\sqrt x - 1}}{{x - 2\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
= \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) - x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}:\dfrac{{2\sqrt x - 1 - \sqrt x + 2}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 2\sqrt x - x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ - 2\sqrt x - 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}}\\
b)A = \dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}} = - \dfrac{{2\left( {\sqrt x + 1} \right) - 2}}{{\sqrt x + 1}}\\
= - 2 + \dfrac{2}{{\sqrt x + 1}}\\
A \in Z \to \dfrac{2}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 2\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
( sửa \(\dfrac{1}{x}\) thành \({\dfrac{1}{{\sqrt x }}}\) bài mới làm đc nha )
