Đáp án:
\(\begin{array}{l}
b)B = \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
c)x = 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
b)B = \dfrac{{x - \sqrt x - 7 + \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 7 + x - 4 - x + 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
c)M = A.B = \dfrac{{x - 3}}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{x - 3}}{{\sqrt x + 3}} = \dfrac{{x - 9 + 6}}{{\sqrt x + 3}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 6}}{{\sqrt x + 3}}\\
= \left( {\sqrt x - 3} \right) + \dfrac{6}{{\sqrt x + 3}}\\
M \in Z \to \dfrac{6}{{\sqrt x + 3}} \in Z\\
\to \sqrt x + 3 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 3 = 6\\
\sqrt x + 3 = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 0\left( l \right)
\end{array} \right.\\
\to x = 9
\end{array}\)
