Toán tìm x thuộc số nguyên , biết a, A=2 trên 2 x+1 b, B=-3 trên 2 x+1 c, C= x+1trên x-2 d,D= 3x-9 trên x-4 e,E= 6x+5 trên 2x-1 giúp mình với 23/07/2021 By Raelynn tìm x thuộc số nguyên , biết a, A=2 trên 2 x+1 b, B=-3 trên 2 x+1 c, C= x+1trên x-2 d,D= 3x-9 trên x-4 e,E= 6x+5 trên 2x-1 giúp mình với
Đáp án: d. \(\left[ \begin{array}{l}x = 7\\x = 1\\x = 5\\x = 3\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.DK:x \ne – \dfrac{1}{2}\\A = \dfrac{2}{{2x + 1}}\\A \in Z \Leftrightarrow \dfrac{2}{{2x + 1}} \in Z\\ \Leftrightarrow 2x + 1 \in U\left( 2 \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x + 1 = 2\\2x + 1 = – 2\\2x + 1 = 1\\2x + 1 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = \dfrac{1}{2}\left( l \right)\\x = – \dfrac{3}{2}\left( l \right)\\x = 0\left( {TM} \right)\\x = – 1\left( {TM} \right)\end{array} \right.\\b.DK:x \ne – \dfrac{1}{2}\\B = – \dfrac{3}{{2x + 1}}\\B \in Z \Leftrightarrow \dfrac{3}{{2x + 1}} \in Z\\ \Leftrightarrow 2x + 1 \in U\left( 3 \right)\\ \to \left[ \begin{array}{l}2x + 1 = 3\\2x + 1 = – 3\\2x + 1 = 1\\2x + 1 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = 1\left( {TM} \right)\\x = – 2\left( {TM} \right)\\x = 0\left( {TM} \right)\\x = – 1\left( {TM} \right)\end{array} \right.\\c.DK:x \ne 2\\C = \dfrac{{x + 1}}{{x – 2}} = \dfrac{{x – 2 + 3}}{{x – 2}} = 1 + \dfrac{3}{{x – 2}}\\C \in Z \Leftrightarrow \dfrac{3}{{x – 2}} \in Z\\ \to x – 2 \in U\left( 3 \right)\\ \to \left[ \begin{array}{l}x – 2 = 3\\x – 2 = – 3\\x – 2 = 1\\x – 2 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = 5\\x = – 1\\x = 3\\x = 1\end{array} \right.\\d.DK:x \ne 4\\D = \dfrac{{3\left( {x – 4} \right) + 3}}{{x – 4}} = 3 + \dfrac{3}{{x – 4}}\\D \in Z \to \dfrac{3}{{x – 4}} \in Z\\ \to x – 4 \in U\left( 3 \right)\\ \to \left[ \begin{array}{l}x – 4 = 3\\x – 4 = – 3\\x – 4 = 1\\x – 4 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = 7\\x = 1\\x = 5\\x = 3\end{array} \right.\\e.DK:x \ne \dfrac{1}{2}\\E = \dfrac{{6x + 5}}{{2x – 1}} = \dfrac{{3\left( {2x – 1} \right) + 8}}{{2x – 1}} = 3 + \dfrac{8}{{2x – 1}}\\E \in Z \Leftrightarrow \dfrac{8}{{2x – 1}} \in Z\\ \Leftrightarrow 2x – 1 \in U\left( 8 \right)\\ \to \left[ \begin{array}{l}2x – 1 = 8\\2x – 1 = – 8\\2x – 1 = 4\\2x – 1 = – 4\\2x – 1 = 2\\2x – 1 = – 2\\2x – 1 = 1\\2x – 1 = – 1\end{array} \right. \to \left[ \begin{array}{l}x = \dfrac{9}{2}\left( l \right)\\x = – \dfrac{7}{2}\left( l \right)\\x = \dfrac{5}{2}\left( l \right)\\x = – \dfrac{3}{2}\left( l \right)\\x = \dfrac{3}{2}\left( l \right)\\x = – \dfrac{1}{2}\left( l \right)\\x = 1\left( {TM} \right)\\x = 0\left( {TM} \right)\end{array} \right.\end{array}\) Trả lời