giup em bai nay
#1
Posted 18-10-2010 - 18:02
$y$= $2x$ - $sqrt{x}$
It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow
#2
Posted 18-10-2010 - 20:01
em giải bằng denta xem!tim gia tri nho nhay cua y voi y>0
$y$= $2x$ - $sqrt{x}$
#3
Posted 18-10-2010 - 20:17
Có thể vẽ đồ thị khảo sát hàm số bậc 2em giải bằng denta xem!
Edited by maths_lovely, 18-10-2010 - 20:42.
#4
Posted 18-10-2010 - 22:06
It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow
#5
Posted 18-10-2010 - 22:12
Edited by PTH_Thái Hà, 19-10-2010 - 22:45.
#6
Posted 18-10-2010 - 23:20
It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow
#7
Posted 19-10-2010 - 16:51
$ 2x - \sqrt x = {\left( {\sqrt {2x} } \right)^2} - 2.\sqrt {2x} .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8} - \dfrac{1}{8} = {\left( {\sqrt {2x} - \dfrac{1}{{2\sqrt 2 }}} \right)^2} - \dfrac{1}{8} \ge - \dfrac{1}{8} $
latex lại lỗi rồi
Anh ! Đề cho y >0 mà ?
@ bboy.: thì e cứ cho cái bậc cao nhất bằng bình phương rồi bậc tiếp theo bằng hai lần cái căn số bậc cao nhất. Dư thừa thiếu tình sau .
Edited by maths_lovely, 19-10-2010 - 16:54.
#8
Posted 19-10-2010 - 17:08
Cho y>0 thì làm gì mà tìm Được?????????$ 2x - \sqrt x = {\left( {\sqrt {2x} } \right)^2} - 2.\sqrt {2x} .\dfrac{1}{{2\sqrt 2 }} + \dfrac{1}{8} - \dfrac{1}{8} = {\left( {\sqrt {2x} - \dfrac{1}{{2\sqrt 2 }}} \right)^2} - \dfrac{1}{8} \ge - \dfrac{1}{8} $
latex lại lỗi rồi
#9
Posted 19-10-2010 - 17:55
It is difficult to say what is impossible, for the dream of yesterday is the hope of today and the reality of tomorrow
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users