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

Nguyễn Mạnh Hùng 6,99999999 >6 nhé

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

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

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