Tìm các số nguyên x để các phân số sau có giá trị là số nguyên:
a, 6/x-5
b, x+5/x+1
c, 2x+1/x-1
d, 6x+5/2x+1
e, 3x+1/2x-5
g, x²+5/x+1
Tìm các số nguyên x để các phân số sau có giá trị là số nguyên:
a, 6/x-5
b, x+5/x+1
c, 2x+1/x-1
d, 6x+5/2x+1
e, 3x+1/2x-5
g, x²+5/x+1
Đáp án:
g) \(\left[ \begin{array}{l}
x = 5\\
x = – 7\\
x = 2\\
x = – 4\\
x = 1\\
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{6}{{x – 5}} \in Z\\
\to x – 5 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x – 5 = 6\\
x – 5 = – 6\\
x – 5 = 3\\
x – 5 = – 3\\
x – 5 = 2\\
x – 5 = – 2\\
x – 5 = 1\\
x – 5 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 11\\
x = – 1\\
x = 8\\
x = 2\\
x = 7\\
x = 3\\
x = 6\\
x = 4
\end{array} \right.\\
b)\dfrac{{x + 5}}{{x + 1}} = \dfrac{{x + 1 + 4}}{{x + 1}} = 1 + \dfrac{4}{{x + 1}}\\
\dfrac{{x + 5}}{{x + 1}} \in Z \to \dfrac{4}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 4\\
x + 1 = – 4\\
x + 1 = 2\\
x + 1 = – 2\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = – 5\\
x = 1\\
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.\\
c)\dfrac{{2x + 1}}{{x – 1}} = \dfrac{{2\left( {x – 1} \right) + 3}}{{x – 1}}\\
= 2 + \dfrac{3}{{x – 1}}\\
\dfrac{{2x + 1}}{{x – 1}} \in Z \to \dfrac{3}{{x – 1}} \in Z\\
\to x – 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x – 1 = 3\\
x – 1 = – 3\\
x – 1 = 1\\
x – 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = – 2\\
x = 2\\
x = 0
\end{array} \right.\\
d)\dfrac{{6x + 5}}{{2x + 1}} = \dfrac{{3\left( {2x + 1} \right) + 2}}{{2x + 1}}\\
= 3 + \dfrac{2}{{2x + 1}}\\
\dfrac{{6x + 5}}{{2x + 1}} \in Z \to \dfrac{2}{{2x + 1}} \in Z\\
\to 2x + 1 \in U\left( 2 \right)\\
\to \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\\
x = – 1
\end{array} \right.\\
e)A = \dfrac{{3x + 1}}{{2x – 5}}\\
\to 2A = \dfrac{{6x + 2}}{{2x – 5}} = \dfrac{{3\left( {2x – 5} \right) + 17}}{{2x – 5}}\\
= 3 + \dfrac{{17}}{{2x – 5}}\\
A \in Z \to \dfrac{{17}}{{2x – 5}} \in Z\\
\to 2x – 5 \in U\left( {17} \right)\\
\to \left[ \begin{array}{l}
2x – 5 = 17\\
2x – 5 = – 17\\
2x – 5 = 1\\
2x – 5 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 11\\
x = – 6\\
x = 3\\
x = 2
\end{array} \right.\\
g)\dfrac{{{x^2} + 5}}{{x + 1}} = \dfrac{{{x^2} – 1 + 6}}{{x + 1}}\\
= \dfrac{{\left( {x – 1} \right)\left( {x + 1} \right) + 6}}{{x + 1}}\\
= \left( {x – 1} \right) + \dfrac{6}{{x + 1}}\\
\dfrac{{{x^2} + 5}}{{x + 1}} \in Z\\
\to \dfrac{6}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 6\\
x + 1 = – 6\\
x + 1 = 3\\
x + 1 = – 3\\
x + 1 = 2\\
x + 1 = – 2\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\\
x = – 7\\
x = 2\\
x = – 4\\
x = 1\\
x = – 3\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)