Minimum
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FacuFeri 10/05/2019 at 11:48
We have : \(\sqrt{xy\left(x-y\right)}=x+y\Leftrightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)
\(xy\left(x-y\right)^2=\dfrac{1}{4}.4xy\left[\left(x+y\right)^2-4xy\right]\le\dfrac{\left(x+y\right)^4}{16}\)
so \(\left(x+y\right)^4\ge16\left(x+y\right)^2\) \(\Leftrightarrow p^4-16p^2\ge0\Leftrightarrow p\ge4\)
Equal sign occurs \(\Leftrightarrow x=2+\sqrt{2};b=2-\sqrt{2}\)
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Huy Toàn 8A (TL) 01/03/2019 at 04:38
\(P=\left(\dfrac{30}{x}+\dfrac{6x}{5}\right)+\left(\dfrac{y}{5}+\dfrac{5}{y}\right)+\dfrac{4}{5}\left(x+y\right)\ge2.\sqrt{\dfrac{30}{x}+\dfrac{6x}{5}}+2.\sqrt{\dfrac{y}{5}+\dfrac{5}{y}}+\dfrac{4}{5}.10\)
\(P=2.6+2.1+8=22\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{30}{x}=\dfrac{6x}{5}\\\dfrac{y}{5}=\dfrac{5}{y}\\x+y=10\end{matrix}\right.=>\left\{{}\begin{matrix}x^2=25\\y^2=25\\x+y=10\end{matrix}\right.\)
\(=>x=y=5\)
\(Pmin=22< =>x=y=5\)
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Huy Toàn 8A (TL) 28/02/2019 at 14:06
\(P=2x+y+\dfrac{30}{x}+\dfrac{5}{y}=\left(x+y\right)\left(\dfrac{5}{x}+\dfrac{5}{y}\right)+\left(\dfrac{25}{x}+x\right)\)
\(\left\{{}\begin{matrix}x>0=>x=\left(\sqrt{x}\right)^2\\y>0=>y=\left(\sqrt{y}\right)^2\end{matrix}\right.\)
\(P1=x+y\ge10\)
\(P2=\dfrac{5}{x}+\dfrac{5}{y}=5.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge5.\dfrac{4}{x+y}=\dfrac{5.4}{10}=2\) khi \(x=y\)
\(P3=\dfrac{25}{x}+x\ge2.\sqrt{\dfrac{25}{x}.x}=2.5=10\) khi \(x=5\)
\(P=\sum P\ge10+2+10=22\) khi \(\left(x;y\right)=\left(5;5\right)\)
Study well
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Vui Ghét Nét 07/03/2019 at 03:53
Reference Lê Anh Duy
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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 Coordinator
05/08/2018 at 03:48
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Tôn Thất Khắc Trịnh 31/07/2018 at 14:11
Mininum 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.
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\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2\ge5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2+2ab+b^2=9\\a^4+2a^2b^2+b^4\ge25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2ab=9-\left(a^2+b^2\right)\\a^4+b^4+2a^2b^2\ge25\end{matrix}\right.\)
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
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There is something is wrong here
Change to :
We have \(a^2+b^2\ge5\)
=> \(9-\left(a^2+b^2\right)\le4\)
=> \(2ab\le4\)
=> \(ab\le2\)
<=> \(a^2b^2\ge4\)
=> \(4a^2b^2\ge16\)
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¤« 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