Maximum
Lê Anh Duy
01/09/2018 at 16:02-
Alone 23/10/2018 at 13:01
\(x^2+5y^2-4xy-x+2y-6=0\)
\(\Leftrightarrow x^2-4xy+4y^2-\left(x-2y\right)-6=-y^2\)
\(\Leftrightarrow\left(x-2y\right)^2-\left(x-2y\right)-6=-y^2\le0\)
\(\Leftrightarrow\left(x-2y+2\right)\left(x-2y-3\right)\le0\)
\(\Leftrightarrow-2\le x-2y\le3\)
Lê Anh Duy selected this answer.
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Searching4You 26/07/2017 at 11:53
Use Cauchy's inequality for positive numbes a,b,c.
\(a+b\ge2\sqrt{ab},b+c\ge2\sqrt{bc},c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{ab}\cdot\sqrt{bc}\cdot\sqrt{ca}=8abc\)
\(\Rightarrow A=\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)
\(MaxA=\dfrac{1}{8}\Leftrightarrow a=b=c>0\)
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¤« 03/04/2018 at 13:32
Use Cauchy's inequality for positive numbes a,b,c.
a+b≥2ab−−√,b+c≥2bc−−√,c+a≥2ca−−√
⇒(a+b)(b+c)(c+a)≥8ab−−√⋅bc−−√⋅ca−−√=8abc
⇒A=abc(a+b)(b+c)(c+a)≤18
MaxA=18⇔a=b=c>0
Summer Clouds moderators
22/07/2017 at 12:01-
We have :
\(x^2-2x+5=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\)
\(\Rightarrow H=\dfrac{1}{x^2-2x+5}\le\dfrac{1}{4}\)
\(\Rightarrow Max_H=\dfrac{1}{4}\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
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¤« 03/04/2018 at 13:33
We have :
x2−2x+5=x2−2x+1+4
=(x−1)2+4≥4
⇒H=1x2−2x+5≤14
⇒MaxH=14
⇔(x−1)2=0
⇔x=1
Dung Trần Thùy
19/03/2017 at 22:06-
¤« 03/04/2018 at 13:33
(*) 1+N=.....=x2/(x-2)2+4
We have x2 ≥
0 , (x-2)2+4 ≥
4 > 0
So 1+N ≥
0 => N ≥
-1 ;equality : x=0
(*)1-N=....=(x-4)2/(x-2)2+4
....... -> N ≤
1 , equality : x=4
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¤« 03/04/2018 at 13:32
There is something is wrong here
Change to :
We have a2+b2≥5
=> 9−(a2+b2)≤4
=> 2ab≤4
=> ab≤2
<=> a2b2≥4
=> 4a2b2≥16
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Phan Văn Hiếu 28/03/2017 at 21:30
trình bày = tiếng việt đc ko tiếng anh ngại viết lắm