Đáp án:
c) \(\left[ \begin{array}{l}
x = 121\\
x = 64\\
x = 4\\
x = 49\\
x = 36\\
x = 16
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 9\\
B = \dfrac{{ - x + 8\sqrt x - 31 - \left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right) + \left( {3\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{ - x + 8\sqrt x - 31 - x + 25 + 3x - 10\sqrt x + 3}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 2\sqrt x - 3}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 5} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 5}}\\
b)B < 1\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 5}} < 1\\
\to \dfrac{{\sqrt x + 1 - \sqrt x + 5}}{{\sqrt x - 5}} < 0\\
\to \dfrac{6}{{\sqrt x - 5}} < 0\\
\to \sqrt x - 5 < 0\\
\to 0 \le x < 25\\
c)B = \dfrac{{\sqrt x + 1}}{{\sqrt x - 5}} = \dfrac{{\sqrt x - 5 + 6}}{{\sqrt x - 5}}\\
= 1 + \dfrac{6}{{\sqrt x - 5}}\\
B \in Z \to \dfrac{6}{{\sqrt x - 5}} \in Z\\
\to \sqrt x - 5 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 5 = 6\\
\sqrt x - 5 = - 6\left( l \right)\\
\sqrt x - 5 = 3\\
\sqrt x - 5 = - 3\\
\sqrt x - 5 = 2\\
\sqrt x - 5 = - 2\\
\sqrt x - 5 = 1\\
\sqrt x - 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 121\\
x = 64\\
x = 4\\
x = 49\\
x = 36\\
x = 16
\end{array} \right.
\end{array}\)
