Toán Lớp 8: Tìm giá trị nguyên của x để P = (4x^2+5)/(2x+1) đạt giá trị nguyên
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Toán Lớp 8: Tìm giá trị nguyên của x để P = (4x^2+5)/(2x+1) đạt giá trị nguyên
Comments ( 2 )
x = 1\\
x = – 2\\
x = 0\\
x = – 1
\end{array} \right.\)
DK:x \ne – \dfrac{1}{2}\\
P = \dfrac{{4{x^2} + 5}}{{2x + 1}}\\
= \dfrac{{4{x^2} – 1 + 6}}{{2x + 1}}\\
= \dfrac{{\left( {2x + 1} \right)\left( {2x – 1} \right) + 6}}{{2x + 1}}\\
= \left( {2x – 1} \right) + \dfrac{6}{{2x + 1}}\\
P \in Z \to \dfrac{6}{{2x + 1}} \in Z\\
\to 2x + 1 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
2x + 1 = 6\\
2x + 1 = – 6\\
2x + 1 = 3\\
2x + 1 = – 3\\
2x + 1 = 2\\
2x + 1 = – 2\\
2x + 1 = 1\\
2x + 1 = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\left( l \right)\\
x = – \dfrac{7}{2}\left( l \right)\\
x = 1\\
x = – 2\\
x = \dfrac{1}{2}\left( l \right)\\
x = – \dfrac{3}{2}\left( l \right)\\
x = 0\\
x = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = – 2\\
x = 0\\
x = – 1
\end{array} \right.
\end{array}\)