Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
M = \frac{{x + \sqrt x - 4}}{{2\sqrt x + 1}}\\
dk:x \ge 0\\
= > 4M = \frac{{4x + 4\sqrt x - 32}}{{2\sqrt x + 1}}\\
= \frac{{4x + 2\sqrt x + 2\sqrt x + 1 - 33}}{{2\sqrt x + 1}}\\
= \frac{{2\sqrt x \left( {2\sqrt x + 1} \right) + \left( {2\sqrt x + 1} \right) - 33}}{{2\sqrt x + 1}}\\
= 2\sqrt x + 1 - \frac{{33}}{{2\sqrt x + 1}}\\
M \in Z = > 4M \in Z \Leftrightarrow \left\{ \begin{array}{l}
2\sqrt x \in Z\\
\frac{{33}}{{2\sqrt x + 1}} \in Z
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = {k^2}\\
2\sqrt x + 1 \in U\left( {33} \right) = \{ \pm 1; \pm 33\}
\end{array} \right.\\
M\`a :2\sqrt x + 1 \ge 1 = > \left[ \begin{array}{l}
2\sqrt x + 1 = 1\\
2\sqrt x + 1 = 33
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2\sqrt x = 0\\
2\sqrt x = 32
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 16
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 256
\end{array} \right.
\end{array}\]

Thử lại

+_ với x=0=> \[M = \frac{{0 + \sqrt 0 - 4}}{{2\sqrt 0 + 1}} = - 4\left( {tm} \right)\]

+) với x=256=> \[M = \frac{{256 + \sqrt {256} - 4}}{{2\sqrt {256} + 1}} = \frac{{268}}{{33}}\left( {ktm} \right)\]

Vậy với x=0 thì M nguyên

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Answer 2

\[\begin{array}{l}
M = \frac{{x + \sqrt x - 4}}{{2\sqrt x + 1}}\\
dk:x \ge 0\\
= > 4M = \frac{{4x + 4\sqrt x - 32}}{{2\sqrt x + 1}}\\
= \frac{{4x + 2\sqrt x + 2\sqrt x + 1 - 33}}{{2\sqrt x + 1}}\\
= \frac{{2\sqrt x \left( {2\sqrt x + 1} \right) + \left( {2\sqrt x + 1} \right) - 33}}{{2\sqrt x + 1}}\\
= 2\sqrt x + 1 - \frac{{33}}{{2\sqrt x + 1}}\\
M \in Z = > 4M \in Z \Leftrightarrow \left\{ \begin{array}{l}
2\sqrt x \in Z\\
\frac{{33}}{{2\sqrt x + 1}} \in Z
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = {k^2}\\
2\sqrt x + 1 \in U\left( {33} \right) = \{ \pm 1; \pm 33\}
\end{array} \right.\\
M\`a :2\sqrt x + 1 \ge 1 = > \left[ \begin{array}{l}
2\sqrt x + 1 = 1\\
2\sqrt x + 1 = 33
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2\sqrt x = 0\\
2\sqrt x = 32
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 16
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 256
\end{array} \right.
\end{array}\]

Thử lại

+_ với x=0=> \[M = \frac{{0 + \sqrt 0 - 4}}{{2\sqrt 0 + 1}} = - 4\left( {tm} \right)\]

+) với x=256=> \[M = \frac{{256 + \sqrt {256} - 4}}{{2\sqrt {256} + 1}} = \frac{{268}}{{33}}\left( {ktm} \right)\]

Vậy với x=0 thì M nguyên

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