Your title is wrong. Here's a example: \(abcd=1.\left(-1\right).1.\left(-1\right)=1\)but          \(a^2+b^2+c^2+d^2+ab+cd=1+1+1+1+\left(-1\right)+\left(-1\right)=2< 6\)

Sorry, \(\dfrac{a+b}{2}.\dfrac{b+c}{2}.\dfrac{c+d}{2}.\dfrac{d+e}{2}.\dfrac{e+a}{2}\ge\dfrac{a+b+c}{3}.\dfrac{b+c+d}{3}.\dfrac{c+d+e}{3}.\dfrac{d+e+a}{3}.\dfrac{e+a+b}{3}\)

We have: \(\left(n^2+n-1\right)^2-1=\left(n^2+n\right)\left(n^2+n-2\right)=\left(n-1\right)n\left(n+1\right)\left(n+2\right)⋮24\left(n\in Z\right)\)                                                                                                                         (product of four consecutive numbers)

So ...

Lê Anh Duy is wrong.

\(6!=1.2.3.4.5.6=9.1.2.4.5.2⋮9\)

So there are: 92+1=93(factorials)

Put : \(f\left(0,n\right)=n+1\)(1)

        \(f\left(k,0\right)=f\left(k-1,1\right)\)(2)

        \(f\left(k+1,n+1\right)=f\left(k,f\left(k+1,n\right)\right)\)(3)

After that we have:

From (3) \(\Rightarrow f\left(2,2\right)=f\left(1,f\left(2,1\right)\right)=f\left(1,f\left(1,f\left(1,1\right)\right)\right)=f\left(1,f\left(1,f\left(0,f\left(1,0\right)\right)\right)\right)\)

From (2) \(\Rightarrow f\left(1,f\left(1,f\left(0,f\left(1,0\right)\right)\right)\right)=f\left(1,f\left(1,f\left(0,f\left(0,1\right)\right)\right)\right)\)

From (1) \(\Rightarrow f\left(1,f\left(1,f\left(0,f\left(0,1\right)\right)\right)\right)=f\left(1,f\left(1,f\left(0,2\right)\right)\right)=f\left(1,f\left(1,3\right)\right)\)

From (3) \(\Rightarrow f\left(1,f\left(1,3\right)\right)=f\left(1,f\left(0,f\left(1,2\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(1,1\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)\)

From (2) \(\Rightarrow f\left(1,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)\)

From (1) \(\Rightarrow f\left(1,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,f\left(0,2\right)\right)\right)\right)=f\left(1,f\left(0,f\left(0,3\right)\right)\right)=f\left(1,f\left(0,4\right)\right)=f\left(1,5\right)\)

From (3) \(\Rightarrow f\left(1,5\right)=f\left(0,f\left(1,4\right)\right)=f\left(0,f\left(0,f\left(1,3\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(1,2\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(1,1\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)\right)\)

From (2) \(\Rightarrow f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(1,0\right)\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)\right)\)

From (1) \(\Rightarrow f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,1\right)\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,f\left(0,2\right)\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,f\left(0,3\right)\right)\right)\right)=f\left(0,f\left(0,f\left(0,4\right)\right)\right)=f\left(0,f\left(0,5\right)\right)=f\left(0,6\right)=7\)

So f(2,2)=7

\(12=\left(6:3\right)!.5+\left(5-3\right)\)

\(12=3.3+3\)

\(12=\dfrac{6.5}{3}+5-3\)

\(12=\dfrac{5!}{6}-5-3\)

\(12=3!+6\)

\(12=3\sqrt{5.3+6-5}\)

\(12=\dfrac{6.6}{3}\)

\(12=\dfrac{6!}{3!}:\left(5+5\right)\)

Change the number 6 in 62, then we have: \(2^6-63=1\left(64-63=1\right)\)

The sum of the digits of m.n is: 8.2014+7+1.2014+2=18135.

\(m.n=\left(999...99\right).\left(888...88\right)\)

                2015            2015

         \(=\left(888...88\right).\left(1000...00-1\right)\)

                2015             2016

        \(=\left(888...88000..00\right)-\left(888...88\right)\)

                     4030                      2015

        \(=888...887111...112\)

               2014         2014

We have: \(S=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

                  \(>\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\right)\)

                                        50                                                      50

                  \(=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)

So \(S>\dfrac{7}{12}\).

We can easily prove that P>0.

We have: \(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}\)

                                                                 \(=1-\dfrac{1}{2013}< 1\)

\(\Rightarrow0< P< 1\)

So P isn't a integer.

We have: \(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^{\text{4}}\)

                    \(=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4\)

                    \(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

                    \(=\left[\left(x^2+5xy+5y^2\right)-y^2\right]\left[\left(x^2+5xy+5y^2\right)+y^2\right]+y^4\)

                    \(=\left(x^2+5xy+5y^2\right)^2-y^4+y^4\)

                    \(=\left(x^2+5xy+5y^2\right)^2\)

So A is a square number.

The difference between the loud rock concert and the normal conversation is 120-60=60 dB. It's 3 times of 20 dB. Because every increase of 20 dB corresponds to sound becoming 10 times as loud, the rock concert must be 10 × 10 × 10 = 1000 times as loud as the normal conversation.

In each day, Gaylon uses 8 digits. We have: 2018:8 = 252 and the remainder is 2. The 253rd day of the year is in September, so Gaylon writes the first two digits of the number is 09 in the month of September. The 2018th digit he writes down is 9.

April has 30 days.

=> Zeus needs to throw: 30.12=360 (lightning bolts) in April.

In the first week of April, he has thrown: 15.7=105 (lightning bolts)

In the next 30-7=23 days, Zeus needs to throw: 360-105=255 (lightning bolts)

So the average is: 255:23=11.08\(\approx\)11 (lightning bolts per day).

Dao Trong Luan is wrong.

When \(0< a,b,c< 1\), \(\sqrt{a}>a;\sqrt{b}>b;\sqrt{c}>c\)

Ex: \(\sqrt{0.36}=0.6>0.36\)

The biggest sum she can get is 24 (19:59)

The smallest positive integer divisible by 2, 3 and 4 is 12.

The number of ways: \(\dfrac{9!}{5!.4!}=126\) (ways)