Đáp án:
b) x=9
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
M = \dfrac{{\sqrt x \left( {\sqrt x + 2} \right) - 6\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 2\sqrt x - 6\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x - 4\sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}}\\
b)M = \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 5}}{{\sqrt x + 2}} = 1 - \dfrac{5}{{\sqrt x + 2}}\\
M \in Z\\
\to \dfrac{5}{{\sqrt x + 2}} \in Z\\
\to \sqrt x + 2 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 5\\
\sqrt x + 2 = 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = - 1\left( l \right)
\end{array} \right.\\
\to x = 9
\end{array}\)
