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
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}\)