![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
![](https://blog.hocmai.vn/wp-content/uploads/2017/07/hot.gif)
<IMG title="s_1 + s_2 = 800 < = > \frac{{a_1 t_1^2 }}{2} + \frac{{a_2 t_2^2 }}{2} = 800 < = > a_1 t_1^2 + a_2 t_2^2 = 1600" alt="s_1 + s_2 = 800 < = > \frac{{a_1 t_1^2 }}{2} + \frac{{a_2 t_2^2 }}{2} = 800 < = > a_1 t_1^2 + a_2 t_2^2 = 1600" src="http://diendan.hocmai.vn/latex.php?s_1 + s_2 = 800 < = > \frac{{a_1 t_1^2 }}{2} + \frac{{a_2 t_2^2 }}{2} = 800 < = > a_1 t_1^2 + a_2 t_2^2 = 1600">
Để t1+t2 min thì a1=a2=a=2m/s2 (đạt max) =>
Theo bất đẳng thức Côsi ta có: <IMG title="t_1^2 + t_2^2 \ge 2t_1 t_2 < = > (t_1 - t_2 )^2 \ge 0" alt="t_1^2 + t_2^2 \ge 2t_1 t_2 < = > (t_1 - t_2 )^2 \ge 0" src=x.php?t_1^2 + t_2^2 \ge 2t_1 t_2 < = > (t_1 - t_2 )^2 \ge 0">
Dấu « = » khi t1=t2=20s => tmin=40s
Để t1+t2 min thì a1=a2=a=2m/s2 (đạt max) =>
Theo bất đẳng thức Côsi ta có: <IMG title="t_1^2 + t_2^2 \ge 2t_1 t_2 < = > (t_1 - t_2 )^2 \ge 0" alt="t_1^2 + t_2^2 \ge 2t_1 t_2 < = > (t_1 - t_2 )^2 \ge 0" src=x.php?t_1^2 + t_2^2 \ge 2t_1 t_2 < = > (t_1 - t_2 )^2 \ge 0">
Dấu « = » khi t1=t2=20s => tmin=40s