Đáp án:
26) x=4
Giải thích các bước giải:
\(\begin{array}{l}
22)D = \dfrac{{x - 6\sqrt x + 9 + 5\sqrt x - 12}}{{\sqrt x - 3}}\\
= \dfrac{{{{\left( {\sqrt x - 3} \right)}^2} + 5\left( {\sqrt x - 3} \right) + 3}}{{\sqrt x - 3}}\\
= \left( {\sqrt x - 3} \right) + 5 + \dfrac{3}{{\sqrt x - 3}}\\
D \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x - 3}} \in Z\\
\Leftrightarrow \sqrt x - 3 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 3\\
\sqrt x - 3 = - 3\\
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 6\\
\sqrt x = 0\\
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 36\\
x = 0\\
x = 16\\
x = 4
\end{array} \right.\\
24)A = \dfrac{{x - 2\sqrt x - 4}}{{\sqrt x - 3}} = \dfrac{{x - 6\sqrt x + 9 + 4\sqrt x - 13}}{{\sqrt x - 3}}\\
= \dfrac{{{{\left( {\sqrt x - 3} \right)}^2} + 4\left( {\sqrt x - 3} \right) - 1}}{{\sqrt x - 3}}\\
= \left( {\sqrt x - 3} \right) + 4 - \dfrac{1}{{\sqrt x - 3}}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x - 3}} \in Z\\
\Leftrightarrow \sqrt x - 3 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 4\\
\sqrt x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 16\\
x = 4
\end{array} \right.\\
26)Q = \dfrac{{x + 10\sqrt x + 25 - 7\sqrt x - 42}}{{\sqrt x + 5}}\\
= \dfrac{{{{\left( {\sqrt x + 5} \right)}^2} - 7\left( {\sqrt x + 5} \right) - 7}}{{\sqrt x + 5}}\\
\left( {\sqrt x + 5} \right) - 7 - \dfrac{7}{{\sqrt x + 5}}\\
Q \in Z\\
\to \dfrac{7}{{\sqrt x + 5}} \in U\left( 7 \right)\\
Mà:\sqrt x + 5 \ge 5\forall x \ge 0\\
\to \sqrt x + 5 = 7\\
\to \sqrt x = 2\\
\to x = 4
\end{array}\)
