Cho x,y>0 thỏa x+y>6
Tìm Min M=$3x+2y+\frac{6}{x}+\frac{8}{y}$
Tìm Min M=$3x+2y+\frac{6}{x}+\frac{8}{y}$
Started By Thang Nguyen2001, 19-10-2015 - 17:33
#1
Posted 19-10-2015 - 17:33
#2
Posted 19-10-2015 - 18:06
$2M=3(x+y)+(3x+\dfrac{12}{x})+(y+\dfrac{16}{y}) \ge 3.6 + 2.2+2.4 = 30$
Đẳng thức xảy ra khi $x=2; y=4$
- Thang Nguyen2001 likes this
#3
Posted 19-10-2015 - 22:49
Cho x,y>0 thỏa x+y>6
Tìm Min M=$3x+2y+\frac{6}{x}+\frac{8}{y}$
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