Đáp án: 2) \(\left[ \begin{array}{l}x = 9\\x = 4\\x = 0\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}1)P = \dfrac{{2x + 4\sqrt x + 6 + \left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) + 3\left( {\sqrt x – 1} \right) – 2\left( {x + 2\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{2x + 4\sqrt x + 6 + x + \sqrt x – 6 + 3\sqrt x – 3 – 2x – 4\sqrt x + 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{x + 4\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}}\\2)P = \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} = \dfrac{{\sqrt x – 1 + 2}}{{\sqrt x – 1}} = 1 + \dfrac{2}{{\sqrt x – 1}}\\P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x – 1}} \in Z\\ \Leftrightarrow \sqrt x – 1 \in U\left( 2 \right)\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x – 1 = 2\\\sqrt x – 1 = – 2\left( l \right)\\\sqrt x – 1 = 1\\\sqrt x – 1 = – 1\end{array} \right. \to \left[ \begin{array}{l}\sqrt x = 3\\\sqrt x = 2\\\sqrt x = 0\end{array} \right.\\ \to \left[ \begin{array}{l}x = 9\\x = 4\\x = 0\end{array} \right.\end{array}\) Reply
Đáp án:
2) \(\left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)P = \dfrac{{2x + 4\sqrt x + 6 + \left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) + 3\left( {\sqrt x – 1} \right) – 2\left( {x + 2\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{2x + 4\sqrt x + 6 + x + \sqrt x – 6 + 3\sqrt x – 3 – 2x – 4\sqrt x + 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{x + 4\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}}\\
2)P = \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} = \dfrac{{\sqrt x – 1 + 2}}{{\sqrt x – 1}} = 1 + \dfrac{2}{{\sqrt x – 1}}\\
P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x – 1}} \in Z\\
\Leftrightarrow \sqrt x – 1 \in U\left( 2 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x – 1 = 2\\
\sqrt x – 1 = – 2\left( l \right)\\
\sqrt x – 1 = 1\\
\sqrt x – 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = 2\\
\sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)