Đáp án:
c. x=4
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne 1\\
E = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + \sqrt x + 2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}:\dfrac{{x - 1 + \sqrt x + 2 - x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}.\dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{x}{{\sqrt x - 1}}\\
b.\left| E \right| = E\\
\to \left[ \begin{array}{l}
E = E\left( {ld} \right)\\
E = - E
\end{array} \right.\\
\to 2E = 0\\
\to E = 0\\
\to \dfrac{x}{{\sqrt x - 1}} = 0\\
\Leftrightarrow x = 0\left( l \right)\\
\to x \in \emptyset \\
c.E = \dfrac{x}{{\sqrt x - 1}} = \dfrac{{x - 1 + 1}}{{\sqrt x - 1}} = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1}}{{\sqrt x - 1}}\\
= \sqrt x + 1 + \dfrac{1}{{\sqrt x - 1}}\\
E \in Z \Leftrightarrow \dfrac{1}{{\sqrt x - 1}} \in Z\\
\Leftrightarrow \sqrt x - 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0\left( l \right)
\end{array} \right.
\end{array}\)
