Đáp án:
\(\begin{array}{l}
a)\dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b)x = 16\\
c)x = 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)M = \left( {1 - \dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 2\sqrt x }}{{x - 1}}\\
= \dfrac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b)M = \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} = \dfrac{1}{2}\\
\to 2\sqrt x - 6 = \sqrt x - 2\\
\to \sqrt x = 4\\
\to x = 16\\
c)M = \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 - 1}}{{\sqrt x - 2}}\\
= 1 - \dfrac{1}{{\sqrt x - 2}}\\
M \in Z \to \dfrac{1}{{\sqrt x - 2}} \in Z\\
\to \sqrt x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 1\left( l \right)
\end{array} \right.\\
\to x = 9
\end{array}\)
