Minimum
Huỳnh Anh Phương
06/11/2018 at 05:26
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.
Quoc Tran Anh Le Coodinator
05/08/2018 at 03:48
Lê Anh Duy
31/07/2018 at 08:58-
Tôn Thất Khắc Trịnh 31/07/2018 at 14:11Mininum divisors? That is the number 1, obviously
Or did you mean the minimum NUMBER of divisors?
If so than it's 9, how you ask?
To create less permutations with the divisors, the factors should overlap the most, hence, the 6 divisor number is a5 and the 4 divisor number is a3, so the product would be a8, having 9 divisors.
Searching4You
26/07/2017 at 12:06-
\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=9\\a^4+2a^2b^2+b^4\ge25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2ab=9-\left(a^2+b^2\right)\\a^4+b^4+2a^2b^2\ge25\end{matrix}\right.\)
We have : a2 + b2 \(\ge\) 5
=> \(-\left(a^2+b^2\right)\le5\)
=> \(9-\left(a^2+b^2\right)\ge9-5=4\)
=> \(2ab\ge4\)
=> \(ab\ge2\)
<=> \(a^2b^2\ge4\)
<=> \(4a^2b^2\ge16\)
Plus \(4a^2b^2\ge16\) into \(a^4+b^4+2a^2b^2\ge25\)
=> \(a^4+b^4+6a^2b^2\ge41\)
=> Min = 41
That is my opinion :v
Selected by MathYouLike -
¤« 03/04/2018 at 13:32{a+b=3a2+b2≥5⇒{a2+2ab+b2=9a4+2a2b2+b4≥25⇒{2ab=9−(a2+b2)a4+b4+2a2b2≥25
We have : a2 + b2 ≥
5
=> −(a2+b2)≤5
=> 9−(a2+b2)≥9−5=4
=> 2ab≥4
=> ab≥2
<=> a2b2≥4
<=> 4a2b2≥16
Plus 4a2b2≥16
into a4+b4+2a2b2≥25
=> a4+b4+6a2b2≥41
=> Min = 41
That is my opinion :v
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
-
There is something is wrong here
Change to :
We have \(a^2+b^2\ge5\)
=> \(9-\left(a^2+b^2\right)\le4\)
=> \(2ab\le4\)
=> \(ab\le2\)
<=> \(a^2b^2\ge4\)
=> \(4a^2b^2\ge16\)
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
-
Phan Văn Hiếu 28/03/2017 at 21:30trình bày = tiếng việt đc ko tiếng anh ngại viết lắm
