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

\(\left(2x_1-5y_1\right)^{2018}\ge0;\left(2x_2-5y_2\right)^{2018}\ge0;.......;\left(2x_{2019}-5y_{2019}\right)^{2018}\ge0\)

\(\Rightarrow\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+......+\left(2x_{2019}-5y_{2019}\right)^{2018}\ge0\)

Because \(\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+..........+\left(2x_{2019}-5y_{2019}\right)^{2018}\le0\)

\(\Rightarrow\left(2x_1-5y_1\right)^{2018}+\left(2x_2-5y_2\right)^{2018}+......+\left(2x_{2019}-5y_{2019}\right)^{2018}=0\)

\(\Rightarrow2x_1=5y_1;2x_2=5y_2;..........;2x_{2019}=5y_{2019}\)

\(\Rightarrow\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=..........=\dfrac{x_{2019}}{y_{2019}}=\dfrac{5}{2}=\dfrac{x_1+x_2+.......+x_{2019}}{y_1+y_2+.......+y_{2019}}\)

\(\left\{{}\begin{matrix}mn=p\\np=m\\mp=n\end{matrix}\right.\)\(\Rightarrow mn.np.mp=p.m.n\Leftrightarrow m^2.n^2.p^2=m.n.p\)

\(\Rightarrow m.n.p.\left(m.n.p-1\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}m.n.p=0\\m.n.p=1\end{matrix}\right.\)

With \(m.n.p=0\Rightarrow p^2=0\Rightarrow p=0\Rightarrow m=n=0\)

With \(m.n.p=1\Rightarrow p^2=1\Rightarrow p=\pm1\)

  If \(p=1\) then \(\left\{{}\begin{matrix}mn=1\\m=n\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}m=n=1\\m=n=-1\end{matrix}\right.\)

  If  \(p=-1\) then \(\left\{{}\begin{matrix}mn=-1\\m=-n\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}m=1,n=-1\\m=-1,n=1\end{matrix}\right.\)

 So have 5 ordered triples of integers satisfied

We have:\(1-\dfrac{1}{1+2+.......+n}=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=1-\dfrac{2}{n\left(n+1\right)}\)

\(=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}=\dfrac{n^2-n+2n-2}{n\left(n+1\right)}=\dfrac{n\left(n-1\right)+2\left(n-1\right)}{n\left(n+1\right)}=\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Apply we have:\(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)..........\left(1-\dfrac{1}{1+2+.......+2006}\right)\)

\(=\dfrac{4.1}{2.3}.\dfrac{5.2}{3.4}........\dfrac{2008.2005}{2006.2007}=\dfrac{4.5.....2008}{2.3.....2006}.\dfrac{1.2......2005}{3.4......2007}=\dfrac{2007.2008}{2.3}.\dfrac{1.2}{2006.2007}\)

\(=\dfrac{2008}{3.2006}=\dfrac{1004}{3.1003}=\dfrac{1004}{3009}\)

We have:\(V_{hinhcau}=3S_{hinhcau}\)

\(\Rightarrow\dfrac{4}{3}.\pi.r^3=3.4.\pi.r^2\) with \(r\) is the radius of this certain

\(\Rightarrow\dfrac{r^3}{r^2}=\dfrac{3.4.\pi}{\dfrac{4}{3}.\pi}=\dfrac{3.4}{\dfrac{4}{3}}=\dfrac{3.4.3}{4}=9\)

\(\Rightarrow r=9\)

We have:\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)

\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2x+2y+2z}{x+y+z}=2\)

\(\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\left(\cdot\right)\\x+z+2=2y\left(\cdot\cdot\right)\\x+y-3=2z\left(\cdot\cdot\cdot\right)\\x+y+z=\dfrac{1}{2}\left(\cdot\cdot\cdot\cdot\right)\end{matrix}\right.\)

\(\left(\cdot\cdot\cdot\cdot\right)-\left(\cdot\right)\Rightarrow x-1=\dfrac{1}{2}-2x\Rightarrow3x=\dfrac{3}{2}\Rightarrow x=\dfrac{1}{2}\)

\(\left(\cdot\cdot\cdot\cdot\right)-\left(\cdot\cdot\right)\Rightarrow y-2=\dfrac{1}{2}-2y\Rightarrow3y=\dfrac{5}{2}\Rightarrow y=\dfrac{5}{6}\)

\(\left(\cdot\cdot\cdot\cdot\right)-\left(\cdot\cdot\cdot\right)\Rightarrow z+3=\dfrac{1}{2}-2z\Rightarrow3z=-\dfrac{5}{2}\Rightarrow z=-\dfrac{5}{6}\)

\(\Rightarrow P=2016x+y^{2017}+z^{2017}=2016.\dfrac{1}{2}+\left(\dfrac{5}{6}\right)^{2017}+\left(-\dfrac{5}{6}\right)^{2017}\)

\(=1008+\left(\dfrac{5}{6}\right)^{2017}-\left(\dfrac{5}{6}\right)^{2017}=1008\)

Call x is the length of the altitude drawn to th hypotenuse of the triangle

The hypotenuse of the triangle is:\(\sqrt{5^2+10^2}=\sqrt{125}\)

The area of the triangle is:\(\dfrac{5.10}{2}=\dfrac{x.\sqrt{125}}{2}\)

\(\Rightarrow5.10=x\sqrt{125}\Rightarrow x=\dfrac{5.10}{\sqrt{125}}=\dfrac{50}{\sqrt{125}}=\dfrac{50}{5\sqrt{5}}=\dfrac{10}{\sqrt{5}}=\sqrt{\dfrac{100}{5}}=\sqrt{20}=2\sqrt{5}\)

Answer:\(2\sqrt{5}\) cm

The edge of the square is:\(\sqrt{8}\)

The diameter of the circle is:\(\sqrt{\left(\sqrt{8}\right)^2+\left(\sqrt{8}\right)^2}=\sqrt{8+8}=\sqrt{16}=4\)

The area of the circle is:\(\dfrac{4}{2}.\dfrac{4}{2}=4\pi\)

The balls of yarn Kim must is:

 750:180=\(\dfrac{25}{6}=4,1\left(66\right)\approx5\)(balls)

  Answer:5 balls

Put the number of age of Chris in 1992 is :a

Put the number of age of Joseph in 1992 is :b

In 1992: b=2a                                                                   (1)

In 1998:b+6=\(\dfrac{3}{2}\left(a+6\right)\)                                                        (2)

Let (2)-(1) we have:6=\(\dfrac{3}{2}\left(a+6\right)-2a\Rightarrow6=\dfrac{3}{2}a+9-2a\)

\(\Rightarrow-3=-\dfrac{1}{2}a\Rightarrow a=-3:\left(-\dfrac{1}{2}\right)\Rightarrow a=6\)

In 2018,the number of age of Chris is:6+2018-1992=32 (ages)

\(a+3\le9\Rightarrow a\le6\Rightarrow a\in\left\{0,1,2,3,4,5,6\right\}\)

Put (a+1)a(a+2)(a+3)=\(n^2\)

\(\Rightarrow1000\left(a+1\right)+100a+10\left(a+2\right)+a+3=n^2\)

\(\Rightarrow1000a+1000+100a+10a+20+a+3=n^2\)

\(\Rightarrow1111a+1023=n^2\)

With \(a=0,a=1,a=2,a=3,a=4,a=5,a=6\) then only a=3 satisfy

Put \(A\left(x\right)=\left(3-4x+x^2\right)^{2016}.\left(3+4x+x^2\right)^{2017}\)

The sum of all coefficients after expanding the expression is:

\(A\left(1\right)=\left(3-4.1+1^2\right)^{2016}.\left(3+4.1+1^2\right)^{2017}\)

\(=0^{2016}.8^{2017}=0\)