Đáp án:
b) x=0
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{{15 - \sqrt x + 2\left( {\sqrt x - 5} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}} \right].\dfrac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \dfrac{{15 - \sqrt x + 2\sqrt x - 10}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}.\dfrac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 5}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x + 5} \right)}}.\dfrac{{\sqrt x - 5}}{{\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x + 1}}\\
b)M = A - B = \dfrac{1}{{\sqrt x + 1}} - \dfrac{{1 - \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 1}}{{\sqrt x + 1}} = 1 - \dfrac{1}{{\sqrt x + 1}}\\
M \in Z \Leftrightarrow \dfrac{1}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 \in U\left( 1 \right)\\
\to \sqrt x + 1 = 1\\
\to x = 0
\end{array}\)
