Toán Q=căn x+1/căn x-3 Tìm x nguyên để Q nguyên 20/09/2021 By Hailey Q=căn x+1/căn x-3 Tìm x nguyên để Q nguyên
\(\begin{array}{l} Q = \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} = \dfrac{{\sqrt x – 3 + 4}}{{\sqrt x – 3}} = 1 + \dfrac{4}{{\sqrt x – 3}}\,\left( {x \ge 0;x \ne 9} \right)\\ Q \in Z \Rightarrow \dfrac{4}{{\sqrt x – 3}} \in Z \Rightarrow \left( {\sqrt x – 3} \right) \in U\left( 4 \right) = \left\{ { – 1;1;4; – 4} \right\}\\ + )\,\sqrt x – 3 = 1 \Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16\,\left( {tm} \right)\\ + )\,\sqrt x – 3 = – 1 \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4\,\left( {tm} \right)\\ + )\,\sqrt x – 3 = 4 \Leftrightarrow \sqrt x = 7 \Leftrightarrow x = 49\,\left( {tm} \right)\\ + )\,\sqrt x – 3 = – 4 \Leftrightarrow \sqrt x = – 1\left( {vn} \right) \end{array}\) Trả lời
\(\begin{array}{l}
Q = \dfrac{{\sqrt x + 1}}{{\sqrt x – 3}} = \dfrac{{\sqrt x – 3 + 4}}{{\sqrt x – 3}} = 1 + \dfrac{4}{{\sqrt x – 3}}\,\left( {x \ge 0;x \ne 9} \right)\\
Q \in Z \Rightarrow \dfrac{4}{{\sqrt x – 3}} \in Z \Rightarrow \left( {\sqrt x – 3} \right) \in U\left( 4 \right) = \left\{ { – 1;1;4; – 4} \right\}\\
+ )\,\sqrt x – 3 = 1 \Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16\,\left( {tm} \right)\\
+ )\,\sqrt x – 3 = – 1 \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4\,\left( {tm} \right)\\
+ )\,\sqrt x – 3 = 4 \Leftrightarrow \sqrt x = 7 \Leftrightarrow x = 49\,\left( {tm} \right)\\
+ )\,\sqrt x – 3 = – 4 \Leftrightarrow \sqrt x = – 1\left( {vn} \right)
\end{array}\)