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

Prove that with all positive real numbers a, b, c, d, e we have:

                        \(\sum\dfrac{a+b}{2}\ge\sum\dfrac{a+b+c}{3}\)

Prove that with all real numbers a, b, c, we have:

            \(6abc+2\sqrt{\prod\left(a^2-ab+b^2\right)}\ge\prod\left(a+b\right)\)

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

Prove that:

  1. \(2\dfrac{5}{9}< \dfrac{5}{3}+\dfrac{8}{3^2}+\dfrac{11}{3^3}+...+\dfrac{302}{3^{100}}< 3\dfrac{1}{2}\)

  2. \(3\dfrac{7}{9}< \dfrac{7}{3}+\dfrac{13}{3^2}+\dfrac{19}{3^3}+...+\dfrac{601}{3^{100}}< 5\)

  3. \(\dfrac{11}{3}+\dfrac{17}{3^2}+\dfrac{23}{3^3}+...+\dfrac{605}{3^{100}}< 7\)

  4. \(\dfrac{4}{3}+\dfrac{13}{3^2}+\dfrac{22}{3^3}+...+\dfrac{904}{3^{101}}< \dfrac{17}{4}\)

  5. \(\dfrac{7}{3}+\dfrac{11}{3^2}+\dfrac{15}{3^3}+...+\dfrac{403}{3^{100}}< 4,5\)

Quickly, please. :v

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)

Jim should receive 3/8 of the proceeds.

How many ordered triples of integers (m, n, p) exist such that mn = p, np = m and mp = n?

What are the maximum number of grape vines that can be planted, not closer than nine feet apart, in a square plot containing one-sixteenth of an acre? 

(Note: each side of this square plot would be 52 feet 2 inches)

Slice the number XII = 12 across the middle to get VII = 7.

In base 2, 1 + 1 = 10

Everyday (starting from question 1) I will post 1 - 2 question(s) to get a point:

- The fastest person who answer correctly will get 1 point/question/answer

- You can still answer my old question at https://e-learning.codienhanoi.edu.vn/discuss/member/leanhduy0206 and get point

THE QUESTION TODAY IS:

3) Find the smallest prime number including 3 digits \(\overline{abc}\) that \(a,\overline{ab},\overline{abc},c,\overline{cb},\overline{cba}\) are also prime numbers

(MYTS - 2018 - Grade 7)

Everyday (starting from question 1) I will post 1 - 2 question(s) to get a point:

- The fastest person who answer correctly will get 1 point/question/answer

- You can still answer my old question at https://e-learning.codienhanoi.edu.vn/discuss/member/leanhduy0206 and get point

THE QUESTION TODAY IS:

11) Mr. Cook bought a big box of his favourite Halloween candies to hand out to trick-or-treaters. However,
he ate half of all the candies himself before the first child came and took her share. He ate half of what
was left before the second child came, and half of what was left again before the third child came and
took all of the remaining candies. If it is known that each child received the same number of candies,
which of the following statements is certainly true about how many candies Mr. Cook originally bought?
(A) It is a multiple of 3

(B) It is a multiple of 4

(C) It is a multiple of 6

(D) It is a multiple of 7

(E) It is a multiple of 11

(Full answer please)

Find the value of  \(P=2016x+y^{2017}+z^{2017}\) when \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\).

Where is "d"?

We have: \(y^2=x\left(x+1\right)\left(x+2\right)\left(x+3\right)\)

           \(\Leftrightarrow y^2+1=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

           \(\Leftrightarrow y^2+1=\left(x^2+3x+1\right)^2\)

\(\Rightarrow y=0\) (Because \(y^2\) and \(\left(x^2+3x+1\right)^2\) is two consecutive square numbers)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\\x=-3\end{matrix}\right.\)

So x=0; x=-1; x=-2; x=-3.