All Questions
Mr. Bee moderators
25/06/2018 at 02:53
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Alone 25/06/2018 at 07:10
\(S^o=\)\(\angle\)BAF + \(\angle\)ABC + \(\angle\)BCD + \(\angle\)CDE + \(\angle\)DEF + \(\angle\)EFA
=\(\angle\)AFB + \(\angle\)BAF + \(\angle\)ABF + \(\angle\)CBF + \(\angle\)BCF + \(\angle\)DCF + \(\angle\)CDF + \(\angle\)EDF + \(\angle\)DEF + \(\angle\)EFD + \(\angle\)CFD + \(\angle\)BFC
=\(180^o+180^o+180^o+180^o=720^o\)
Selected by MathYouLike -
Sory, I mistyped. It should've been \(720^o\) not \(540^o.\)
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huynh anh phuong 02/08/2019 at 14:01
OMG !I don't khow what do you say!
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A.R.M.Y 09/08/2018 at 15:55
I don't know I'm sorry but i'm grade 4 =)
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Nguyễn Mạnh Hùng 13/06/2018 at 09:35
We have:
\(a^3+b^3=\left(a+b\right)\left(a^2+ab+b^2\right)=2\) (*)
Other way:
\(a^2+ab+b^2=a^2+2.\dfrac{1}{2}b+\left(\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)
\(=\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\)
Because \(\left(a+\dfrac{1}{2}b\right)^2\ge0\) with \(\forall a,b\)
\(\dfrac{3}{4}b^2\ge0\) with \(\forall b\)
So \(\left(a+\dfrac{1}{2}b\right)^2+\dfrac{3}{4}b^2\ge0\) with \(\forall a,b\)
or \(a^2+ab+b^2\ge0\) with \(\forall a,b\) (**)
From (*) and (**) we have \(a+b\le2\)
Your ex is so hard to do :)
Lê Quốc Trần Anh selected this answer. -
Lê Thành 23/06/2018 at 05:27
We have:
a3+b3=(a+b)(a2+ab+b2)=2a3+b3=(a+b)(a2+ab+b2)=2 (*)
Other way:
a2+ab+b2=a2+2.12b+(12b)2+34b2a2+ab+b2=a2+2.12b+(12b)2+34b2
=(a+12b)2+34b2=(a+12b)2+34b2
Because (a+12b)2≥0(a+12b)2≥0 with ∀a,b∀a,b
34b2≥034b2≥0 with ∀b∀b
So (a+12b)2+34b2≥0(a+12b)2+34b2≥0 with ∀a,b∀a,b
or a2+ab+b2≥0a2+ab+b2≥0 with ∀a,b∀a,b (**)
From (*) and (**) we have a+b≤2a+b≤2
Your ex is so hard to do :)
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Alone 13/06/2018 at 10:16
With x=2 then \(4P\left(2\right)=5.2^2=20\Rightarrow P\left(2\right)=5\)
With x=3 then \(P\left(3\right)+3P\left(2\right)=5.3^2=45\Rightarrow P\left(3\right)=45-5.3=30\)
Lê Quốc Trần Anh selected this answer.
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Alone 14/06/2018 at 01:07
\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{zx}{80}=\dfrac{xy+yz+zx}{24+30+80}=\dfrac{1206}{134}=9\)
\(\Rightarrow x^2=64.9=8^2.3^2\Rightarrow x=\pm24\)
With x=24 then y=24:8.3=9;z=30
With x=-24 then y=-9;z=-30
Lê Quốc Trần Anh selected this answer. -
Nguyễn Mạnh Hùng 15/06/2018 at 01:35
We have:
\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\)
\(\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}\)
Apply the same sequence properties, we have:
\(\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{xz}{80}=\dfrac{xy+yz+xz}{24+30+80}=\dfrac{1206}{134}=9\)
So \(x^2=9.64=576\)\(\Rightarrow x=24\)
\(\dfrac{xy}{24}=9\Rightarrow xy=24.9=216\Rightarrow y=9\)
\(\dfrac{yz}{30}=9\Rightarrow yz=270\Rightarrow z=30\)
So \(\left(x;y;z\right)=\left(24;9;30\right)\)