Đáp án:
\(\left[ \begin{array}{l}
x = 256\\
x = 121\\
x = 1\\
x = 64\\
x = 16\\
x = 49\\
x = 25
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 1;x \ne 36\\
A = \dfrac{{\sqrt x + 4}}{{\sqrt x - 1}}\\
N = \dfrac{{\sqrt x - 6}}{{\sqrt x - 1}}\\
P = \dfrac{A}{N} = \dfrac{{\sqrt x + 4}}{{\sqrt x - 1}}:\dfrac{{\sqrt x - 6}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x - 1}}.\dfrac{{\sqrt x - 1}}{{\sqrt x - 6}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x - 6}} = \dfrac{{\sqrt x - 6 + 10}}{{\sqrt x - 6}} = 1 + \dfrac{{10}}{{\sqrt x - 6}}\\
Để:P \in Z\\
\to \dfrac{{10}}{{\sqrt x - 6}} \in Z\\
\to \sqrt x - 6 \in U\left( {10} \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 6 = 10\\
\sqrt x - 6 = 5\\
\sqrt x - 6 = - 5\\
\sqrt x - 6 = 2\\
\sqrt x - 6 = - 2\\
\sqrt x - 6 = 1\\
\sqrt x - 6 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = {16^2}\\
x = {11^2}\\
x = 1\\
x = {8^2}\\
x = {4^2}\\
x = {7^2}\\
x = {5^2}
\end{array} \right. \to \left[ \begin{array}{l}
x = 256\\
x = 121\\
x = 1\\
x = 64\\
x = 16\\
x = 49\\
x = 25
\end{array} \right.
\end{array}\)
