Tìm số thực x để 5x/x^2 – 4 đạt giá trị nguyên 18/11/2021 Bởi Audrey Tìm số thực x để 5x/x^2 – 4 đạt giá trị nguyên
Đáp án: \(\left[ \begin{array}{l}x = 3\\x = – 3\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}A = \dfrac{{5x}}{{{x^2} – 4}} = \dfrac{{5\left( {x – 2} \right) + 10}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{5}{{x + 2}} + \dfrac{{10}}{{{x^2} – 4}}\\A \in Z \Leftrightarrow \left\{ \begin{array}{l}\dfrac{5}{{x + 2}} \in Z\\\dfrac{{10}}{{{x^2} – 4}} \in Z\end{array} \right.\\ \to \left\{ \begin{array}{l}x + 2 \in U\left( 5 \right)\\{x^2} – 4 \in U\left( {10} \right)\end{array} \right.\\ \to \left\{ \begin{array}{l}\left[ \begin{array}{l}x + 2 = 5\\x + 2 = – 5\\x + 2 = 1\\x + 2 = – 1\end{array} \right.\\\left[ \begin{array}{l}{x^2} – 4 = 10\\{x^2} – 4 = – 10\left( l \right)\\{x^2} – 4 = 5\\{x^2} – 4 = – 5\left( l \right)\\{x^2} – 4 = 2\\{x^2} – 4 = – 2\\{x^2} – 4 = 1\\{x^2} – 4 = – 1\end{array} \right.\end{array} \right. \to \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 3\\x = – 7\\x = – 1\\x = – 3\end{array} \right.\\\left[ \begin{array}{l}x = \pm \sqrt {14} \\x = 3\\x = – 3\\x = \pm \sqrt 6 \\x = \pm \sqrt 2 \\x = \pm \sqrt 5 \\x = \pm \sqrt 3 \end{array} \right.\end{array} \right.\\ \to \left[ \begin{array}{l}x = 3\\x = – 3\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x = 3\\
x = – 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{5x}}{{{x^2} – 4}} = \dfrac{{5\left( {x – 2} \right) + 10}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{5}{{x + 2}} + \dfrac{{10}}{{{x^2} – 4}}\\
A \in Z \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{5}{{x + 2}} \in Z\\
\dfrac{{10}}{{{x^2} – 4}} \in Z
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + 2 \in U\left( 5 \right)\\
{x^2} – 4 \in U\left( {10} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x + 2 = 5\\
x + 2 = – 5\\
x + 2 = 1\\
x + 2 = – 1
\end{array} \right.\\
\left[ \begin{array}{l}
{x^2} – 4 = 10\\
{x^2} – 4 = – 10\left( l \right)\\
{x^2} – 4 = 5\\
{x^2} – 4 = – 5\left( l \right)\\
{x^2} – 4 = 2\\
{x^2} – 4 = – 2\\
{x^2} – 4 = 1\\
{x^2} – 4 = – 1
\end{array} \right.
\end{array} \right. \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 3\\
x = – 7\\
x = – 1\\
x = – 3
\end{array} \right.\\
\left[ \begin{array}{l}
x = \pm \sqrt {14} \\
x = 3\\
x = – 3\\
x = \pm \sqrt 6 \\
x = \pm \sqrt 2 \\
x = \pm \sqrt 5 \\
x = \pm \sqrt 3
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – 3
\end{array} \right.
\end{array}\)