cho M= ( căn x + 6)/(căn x + 1). tìm x để m thuộc z 20/09/2021 Bởi Claire cho M= ( căn x + 6)/(căn x + 1). tìm x để m thuộc z
\(\begin{array}{l} M = \frac{{\sqrt x + 6}}{{\sqrt x + 1}} = 1 + \frac{5}{{\sqrt x + 1}}\\ Do\,\,\sqrt x \ge 0 \Rightarrow \sqrt x + 1 > 0 \Rightarrow M > 1\\ Do\,\,\sqrt x + 1 \ge 1 \Leftrightarrow \frac{5}{{\sqrt x + 1}} \le 5 \Rightarrow M \le 6\\ Suy\,ra\,1 < M \le 6 \Rightarrow M \in \left\{ {2;3;4;5;6} \right\}\\ + )M = 2 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 2 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 1 \Leftrightarrow x = 16\left( {tm} \right)\\ + )M = 3 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 3 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 2 \Leftrightarrow x = \frac{9}{4}\left( {tm} \right)\\ + )M = 4 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 4 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 3 \Leftrightarrow x = \frac{4}{9}\left( {tm} \right)\\ + )M = 5 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 5 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 4 \Leftrightarrow x = \frac{1}{{16}}\left( {tm} \right)\\ + )M = 6 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 6 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 5 \Leftrightarrow x = 0\left( {tm} \right) \end{array}\) Bình luận
\(\begin{array}{l}
M = \frac{{\sqrt x + 6}}{{\sqrt x + 1}} = 1 + \frac{5}{{\sqrt x + 1}}\\
Do\,\,\sqrt x \ge 0 \Rightarrow \sqrt x + 1 > 0 \Rightarrow M > 1\\
Do\,\,\sqrt x + 1 \ge 1 \Leftrightarrow \frac{5}{{\sqrt x + 1}} \le 5 \Rightarrow M \le 6\\
Suy\,ra\,1 < M \le 6 \Rightarrow M \in \left\{ {2;3;4;5;6} \right\}\\ + )M = 2 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 2 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 1 \Leftrightarrow x = 16\left( {tm} \right)\\ + )M = 3 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 3 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 2 \Leftrightarrow x = \frac{9}{4}\left( {tm} \right)\\ + )M = 4 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 4 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 3 \Leftrightarrow x = \frac{4}{9}\left( {tm} \right)\\ + )M = 5 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 5 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 4 \Leftrightarrow x = \frac{1}{{16}}\left( {tm} \right)\\ + )M = 6 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 6 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 5 \Leftrightarrow x = 0\left( {tm} \right) \end{array}\)