## Làm giúp mình bài 1 với

Question

Làm giúp mình bài 1 với

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8 tháng 2020-11-02T12:20:17+00:00 1 Answers 56 views 0

2) $$\left[ \begin{array}{l} x = 9\\ x = 4\\ x = 0 \end{array} \right.$$
$$\begin{array}{l} 1)P = \dfrac{{2x + 4\sqrt x + 6 + \left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) + 3\left( {\sqrt x – 1} \right) – 2\left( {x + 2\sqrt x – 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{2x + 4\sqrt x + 6 + x + \sqrt x – 6 + 3\sqrt x – 3 – 2x – 4\sqrt x + 6}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{x + 4\sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}\\ = \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}}\\ 2)P = \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} = \dfrac{{\sqrt x – 1 + 2}}{{\sqrt x – 1}} = 1 + \dfrac{2}{{\sqrt x – 1}}\\ P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x – 1}} \in Z\\ \Leftrightarrow \sqrt x – 1 \in U\left( 2 \right)\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt x – 1 = 2\\ \sqrt x – 1 = – 2\left( l \right)\\ \sqrt x – 1 = 1\\ \sqrt x – 1 = – 1 \end{array} \right. \to \left[ \begin{array}{l} \sqrt x = 3\\ \sqrt x = 2\\ \sqrt x = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = 9\\ x = 4\\ x = 0 \end{array} \right. \end{array}$$