The speed limit in this zone in feet per second is:
    88:60 x 15==22(feet)

\(x^2+5y^2-4xy-x+2y-6=0\)

\(\Leftrightarrow x^2-4xy+4y^2-\left(x-2y\right)-6=-y^2\)

\(\Leftrightarrow\left(x-2y\right)^2-\left(x-2y\right)-6=-y^2\le0\)

\(\Leftrightarrow\left(x-2y+2\right)\left(x-2y-3\right)\le0\)

\(\Leftrightarrow-2\le x-2y\le3\)

Because  \(\dfrac{12x}{25}\) is a square number but 12=2².3

So \(x⋮25\) and x smallest so x=3.25=75

\(2=\dfrac{1}{5}\Rightarrow2:2.7=\dfrac{1}{5}:2.7\)

\(\Rightarrow7=\dfrac{7}{10}\)

f(3)=3.3-7=9-7=2

\(\Rightarrow g\left(f\left(3\right)\right)=g\left(2\right)=2^2-4=0\)

\(\Rightarrow f\left(g\left(f\left(3\right)\right)\right)=f\left(0\right)=3.0-7=-7\)

\(x^3+y^3+z^3=3xyz\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+z=0\\x^2+y^2+z^2-xy-yz-zx=0\end{matrix}\right.\) Because x,y,z>0 so x+y+z>0

\(\Rightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\)\(\Rightarrow P=\dfrac{x^{2018}+1}{x^{2018}+x^{2018}+x^{2018}+3}=\dfrac{x^{2018}+1}{3.x^{2018}+3}=\dfrac{1}{3}\)

We have:\(9x^2+4y^2=20xy\Leftrightarrow9x^2-20xy+4y^2=0\)

\(\Leftrightarrow9x^2-18xy-2xy+4y^2=0\Leftrightarrow9x\left(x-2y\right)-2y\left(x-2y\right)=0\)

\(\Leftrightarrow\left(9x-2y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}9x=2y\\x=2y\end{matrix}\right.\).Because x<2y \(\Rightarrow9x=2y\Rightarrow P=\dfrac{3x-9x}{3x+9x}=\dfrac{-6x}{12x}=-\dfrac{1}{2}\)

\(S^o=\)\(\angle\)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

 =\(180^o+180^o+180^o+180^o=720^o\)

Draw AK is the altitude of triangle ABC(\(K\in BC\)),the altitude of triangle ADC from point C cut AD at I

So \(BK=CK=\dfrac{1}{2}BC\)

Applying Pitago's theorem we have:

 \(AK^2+CK^2=AC^2\Rightarrow AK^2=AC^2-CK^2=AC^2-\left(\dfrac{1}{2}BC\right)^2=6^2-3^2=27\)

\(\Rightarrow AK=3\sqrt{3}\)\(\Rightarrow S_{ADC}=\dfrac{AK.CD}{2}=\dfrac{3\sqrt{3}.4}{2}=6\sqrt{3}\)

Because BD+DK=BK \(\Rightarrow DK=BK-BD=3-2=1\)

Applying Pitago's theorem in triangle ADK we have:

 \(AD^2=AK^2+DK^2=27+1^2=28\)\(\Rightarrow AD=2\sqrt{7}\)

But \(S_{ADC}=6\sqrt{3}\)\(\Rightarrow\dfrac{AD.CH}{2}=6\sqrt{3}\Rightarrow AD.CH=12\sqrt{3}\Rightarrow CH=\dfrac{12\sqrt{3}}{2\sqrt{7}}=\dfrac{6\sqrt{21}}{7}\)

Because x,y integer so x+y integer so x+y \(\le39\)

We have:\(\left(x-y\right)^2\ge0\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow x^2+2xy+y^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}\le\dfrac{38^2}{4}=361\)(Because \(39^2⋮̸4\))

So maximum of xy is 361 when x=y=19

\(\dfrac{x}{8}=\dfrac{y}{3}=\dfrac{z}{10}\Rightarrow\dfrac{x^2}{64}=\dfrac{xy}{24}=\dfrac{yz}{30}=\dfrac{zx}{80}=\dfrac{xy+yz+zx}{24+30+80}=\dfrac{1206}{134}=9\)

\(\Rightarrow x^2=64.9=8^2.3^2\Rightarrow x=\pm24\)

With x=24 then y=24:8.3=9;z=30

With x=-24 then y=-9;z=-30

With x=0 then \(0.f\left(2018\right)=2016.f\left(0\right)\Rightarrow f\left(0\right)=0\)

With x=-2018 then \(-2018.f\left(-2018+2018\right)=\left(-2018+2016\right).f\left(-2018\right)\)

\(\Rightarrow-2018.f\left(0\right)=-2.f\left(-2018\right)\).Because \(f\left(0\right)=0\) so \(-2f\left(-2018\right)=0\Rightarrow f\left(-2018\right)=0\)

So \(f\left(x\right)=0\) with x=0 and -2018

So 

With x=2 then \(4P\left(2\right)=5.2^2=20\Rightarrow P\left(2\right)=5\)

With x=3 then \(P\left(3\right)+3P\left(2\right)=5.3^2=45\Rightarrow P\left(3\right)=45-5.3=30\)

We have:\(a+\dfrac{1}{b}+b+\dfrac{1}{c}+c+\dfrac{1}{a}< 6\)

But \(a+\dfrac{1}{a}+b+\dfrac{1}{b}+c+\dfrac{1}{c}\ge2\sqrt{a.\dfrac{1}{a}}+2\sqrt{b.\dfrac{1}{b}}+2\sqrt{c.\dfrac{1}{c}}=2+2+2=6\)

So there are no a,b,c>0 satisfys

We have:\(x:y:z=a:b:c\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)

\(\Rightarrow\left(x+y+z\right)^2=\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)

We have:\(\left(ac+bd\right).\left(ad+bc\right)=0\)

\(\Leftrightarrow a^2cd+bad^2+abc^2+b^2cd=0\)

\(\Leftrightarrow\left(a^2+b^2\right).cd+ba\left(c^2+d^2\right)=0\)

\(\Leftrightarrow2017.ab+2017.cd=0\Leftrightarrow ab+cd=0\)

We have:5=3+2

Because \(3⋮3\) but \(2⋮̸3\) so the answer is B.2

\(x^4-8x^2-9=0\Leftrightarrow x^4+x^2-9x^2-9=0\)

\(\Leftrightarrow x^2\left(x^2+1\right)-9\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2-9\right)\left(x^2+1\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+1\right)=0\)

Because \(x^2+1\ne0\) so \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

With n=0 then \(2^0+1=1+1=2⋮̸3\)(not satisfy)

With n=1 then \(2^{2.1}+1=2+1=3⋮3\)(satisfy)

With n>1 then \(2^{2n}=\left(2^2\right)^n=4^n\equiv1^n=1\)(mod 3)\(\Rightarrow2^{2n}+1\equiv2\)(mod 3)

\(\Rightarrow2^{2n}+1⋮̸3\)(not satisfy)

Answer: n=1 

oh;the first row :x,y,z not a,y,z

Because the role of a,y,z is the same so we suppose \(x\le y\le z\)

\(2^z< 2336\Rightarrow z\le11\)

We have:\(2^x+2^y+2^z\le3.2^z\Leftrightarrow2336\le3.2^z\)\(\Leftrightarrow2^z\ge\dfrac{2336}{3}\Rightarrow z\ge10\)

So z\(\in\left\{10,11\right\}\)

With z=10 then \(2^x+2^y=2336-2^{10}=1312\)

    \(2^x+2^y\le2.2^y\Rightarrow2.2^y\ge1312\Rightarrow2^y\ge656\Rightarrow y\ge10\)

  \(2^y< 1312\Rightarrow y\le10\)

 So y=10 so \(2^x=2336-2^{10}-2^{10}=288\)(not satisfy)

With z=11 then \(2^x+2^y=288\)

 \(2^y< 288\Rightarrow y\le8\)

 \(2^y.2\ge2^x+2^y=288\Rightarrow2^y\ge144\Rightarrow y\ge8\)

So y=8 then \(2^x=288-2^8=32\Rightarrow x=5\)

Answer :(x,y,z) is the permutation of \(\left(5,8,11\right)\)

To try on: n=1 then \(A=1^3=1;\dfrac{1^2\left(1+1\right)^2}{4}=\dfrac{4}{4}=1\)

Suppose A be true for n = k; that is:

 \(1^3+2^3+......+k^3=\dfrac{k^2\left(k+1\right)^2}{4}\)

We shall show that A is true too

Add \(\left(k+1\right)^3\),we have:

 \(1^3+2^3+........+k^3+\left(k+1\right)^3\)=\(\dfrac{k^2\left(k+1\right)^2}{4}+\left(k+1\right)^3=\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}\)

\(=\dfrac{\left(k+1\right)^2.\left(k^2+4k+4\right)}{4}=\dfrac{\left(k+1\right)^2.\left(k+2\right)^2}{4}\)

So the supposition right

a,The value of A=\(2\times\left|2008-2007\right|+\left|2008-2009\right|\)\(=\)\(2\times\left|1\right|+\left|-1\right|=2\times1+1=2+1=3\)

b,With \(x< 2007\)then A=x-2010

\(\Rightarrow2\times-\left(x-2007\right)-\left(x-2009\right)=x-2010\)

\(\Leftrightarrow-2x+4014-x+2009=x-2010\)\(\Leftrightarrow-4x=-2010-4014-2009\Leftrightarrow-4x=-8033\)

\(\Leftrightarrow x=\dfrac{8033}{4}>\dfrac{8028}{4}=2007\)(not satisfy)

With \(2007\le x< 2009\) then \(2\times\left(x-2007\right)-\left(x-2009\right)=x-2010\)

\(\Leftrightarrow2x-4014-x+2009=x-2010\)

\(\Leftrightarrow2x-x-x=-2010-2009+4014\)\(\Leftrightarrow0=-5\)(not satisfy)

With \(x\ge2009\) then 

\(2\times\left(x-2007\right)+\left(x-2009\right)=x-2010\)

\(\Leftrightarrow2x-4014+x-2009=x-2010\)

\(\Leftrightarrow2x=-2010+2009+4014\Leftrightarrow2x=4013\)

\(\Leftrightarrow x=2006,5< 2009\)(not satisfy)

 So not have x satisfy

c,We have:A=\(\left|x-2007\right|+\left|x-2009\right|+\left|x-2007\right|\)

\(=\left|x-2007\right|+\left|2009-x\right|+\left|x-2007\right|\)\(\ge\left|x-2007+2009-x\right|+\left|x-2007\right|=\left|2\right|+\left|x-2007\right|\ge2\)

So the smallest of A is 2 when \(x=2007\)

Suppose \(\sqrt{2}\) can be expressed in the form \(\dfrac{m}{n}\) with (m,n)=1

\(\Rightarrow\dfrac{m}{n}=\sqrt{2}\Rightarrow\dfrac{m^2}{n^2}=2\)\(\Rightarrow m^2=2n^2\)

Because \(2n^2⋮2\) so \(m^2⋮2\Rightarrow m⋮2\)(1)

\(\Rightarrow m^2⋮4\)\(\Rightarrow n^2⋮2\Rightarrow n⋮2\)(2)

From (1) and (2),we have the assumption is wrong

So \(\sqrt{2}\) can be expressed in the form \(\dfrac{m}{n}\)

\(f\left(1\right)=f\left(0\right)+1=1+1=2\)

\(f\left(2\right)=f\left(1\right)+1=2+1=3\)

\(f\left(3\right)=f\left(2\right)+1=3+1=4\)

...

So \(f\left(x\right)=x+1\)

\(1000a+100b+10c+d+100b+10c+d+10c+d+d=2018\)

\(\Rightarrow1000a+200b+30c+4d=2018\)

Because 1000a\(\le2018\)\(\Rightarrow a\le2\) and a\(\ne0\Rightarrow a\in\left\{1,2\right\}\)

With a=1 then \(200b+30c+4d=1018\)

  The same \(b\in\left\{0,1,2,3,4,5\right\}\)

   +,b=0\(\Rightarrow30c+4d=1018\).Because \(30c+4d\le30.9+4.9=306\) then the expression isn't satisfy

  +,b=1\(\Rightarrow30c+4d=818\)(not satisfy)

  +,b=2\(\Rightarrow30c+4d=618\)(not satisfy)

  +,b=3\(\Rightarrow30c+4d=418\)(not satisfy)

 +,b=4\(\Rightarrow30c+4d=218\)

    The same:c\(\in\left\{0,1,2,3,4,5,6,7\right\}\).Because \(4d\le36\Rightarrow218-4d\ge218-36=182\)

               \(\Rightarrow30c\ge182\Rightarrow c\in\left\{6,7\right\}\)

             If c=6 then 4d=38\(\Rightarrow d=\dfrac{19}{2}\)(not satisfy)

             If c=7 then 4d=8\(\Rightarrow d=2\)(satisfy)

  +,b=5\(\Rightarrow30c+4d=18\)(not satisfy)

With a=2 then 200b+30c+4d=18(not satisfy)

Answer:The old of Le Quoc Tran Anh is:2018-1472=546(years)

OMG,really,miraculous

Solution 1:\(\sqrt{1+204.205.206.207}=\sqrt{1783288441}=42229\)

Solution 2:\(\sqrt{1+n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\sqrt{1+\left(n^2+3n\right)\left(n^2+3n+2\right)}=\sqrt{\left(n^2+3n\right)^2+2.\left(n^2+3n\right)+1}\)

\(=\sqrt{\left(n^2+3n+1\right)^2}=n^2+3n+1\)

With n=204 then \(\sqrt{1+204.205.206.207}=204^2+3.204+1=42229\)

\(A=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-6^2=\left(x^2+5x\right)^2-36\ge-36\)

Then the minimum of A=-36

Prove that \(\dfrac{n}{\sqrt{n-1}}\ge2\) with \(n>1\)

Prove that P\(=\dfrac{1}{2^2}+\dfrac{1}{3^2}+.......+\dfrac{1}{2013^2}\) isn't a integer

Prove that A=(x+y)(x+2y)(x+3y)(x+4y)+y⁴ is a square of a number