Sorry, can I answer again?

\(\left(x+2\right)\left|x+3\right|=0\)

=> \(\left(x+2\right)=0\) or \(\left|x+3=0\right|\)

=> \(x=\left\{-2;-3;-6\right\}\)

So the answer is \(x=\left\{-2;-3;-6\right\}\) (Phan Thanh Tinh miss one answer)

\(\left(x+2\right)\left|x+3\right|=0\)

=> \(\left(x+2\right)=0\) or \(\left|x+3\right|=0\)

=> \(x=\left\{-2;-3;-6\right\}\)

So there are 3 solutions for this question.

1st solution: \(x=-2\) => The negative root of \(x\) is \(x=2\)

=> \(\left(x+2\right)\left|x+3\right|=\left(2+2\right)\left|2+3\right|=4.5=20\).

2nd solution: \(x=-3\) => The negative root of \(x\) is \(x=3\)

=> \(\left(x+2\right)\left|x+3\right|=\left(3+2\right)\left|3+3\right|=5.6=30\)

3rd solution: \(x=-6\) => The negative root of \(x\) is \(x=6\).

=> \(\left(x+2\right)\left|x+3\right|=\left(6+2\right)\left|6+3\right|=8.9=72\)

So the answer is: \(20;30;72\)

\(x\in\left\{0;3;6;9;...;33;36;39\right\}\)

\(2^{51}-1=2^{3.17}-1=8^{17}-1=\left(8-1\right).\left[\left(8^{17-1}\right)+\left(8^{17-2}\right)+...+\left(8^{17-16}\right)+8^{17-17}\right]\) is divisible by 7.

The question must be changed to: "How much does the small boards and big boards cost all".

They all bought: \(5+3+1+2+4+6=21\left(big\right)\) boards.

They all bought: \(2+4+6+5+3+1=21\left(small\right)\) boards, too.

So it all cost: \(21.\left(486000+88000\right)=12054000\left(dollars\right)\).

So the answer is: $12.054.000

Sorry, but what is your question?

There are total: \(25.2=50\left(shoes\right)\).

Because it needs one shoe for each of it's 100 feet, than the centipede must buy:

\(100-50=50\left(shoes\right)\) or \(25\) pairs of shoes.

So the answer is \(50\left(shoes\right)\) or \(25\) pairs of shoes.

If \(a,b,c\in N\) then I can answer it.

\(a+b+c=0\)

=> \(a=b=c=0\).

Because \(\forall x\) in the case \(0^x=0\) and a number multiplies by \(0\) equals to \(0\).

=> \(a^3+b^3+c^3=3abc=0\left(proved\right)\)

Goku, don't copy Phan Thanh Tinh's answer.

Searching4You, the team that can do all this are call "moderators". They manage Mathu.

I agree with Phan Thanh Tinh, they copy our solution and they get more points. 

There are so many members now had broken the rules so much so that I agree with Phan Thanh Tinh.

I had suggested the management (Summer Clouds) about these but she can only reduce the spam question.

The area of the rectangle is:

\(7.2=14\left(m^2\right)\)

So the answer is \(14m^2\)

Sorry, can I add some more?

If \(\left|x-1\right|+\left|x-3\right|\) has the smallest value then \(\left|x-1\right|or\left|x+3\right|\) must be \(0\left(\left|x-1\right|and\left|x-3\right|\in N\right)\). Then the smallest value is: \(0+2=2\)

The smallest value of \(\left|x-1\right|and\left|x-3\right|\) is \(\left|x-1\right|-\left|x-3\right|=2\).

So the answer is \(2\).

There are \(9\) digits could be the ones digits: \(0,1,2,3,4,5,6,7,8,9.\)

If the ones digit is \(0\) then there are: \(9-0=9\left(numbers\right)\) satisfy the question.

If the ones digit is \(1\) then there are: \(9-1=8\left(numbers\right)\) satisfy the question.

...............................................................................................................................

If the ones digit is \(9\) then there are: \(9-9=0\left(numbers\right)\) satisfy the question.

So there are total: \(9+8+7+6+5+4+3+2+1+0=45\left(numbers\right)\) satisfy the question.

\(A=\dfrac{\left(2^2\right)^{20}.\left(3^2\right)^{25}.\left(5^2\right)^4.\left(3^2\right)}{4^{21}.9^{25}.3^2.5^8}\)

\(A=\dfrac{4^{20}.9^{25}.5^8.3^2}{4^{21}.9^{25}.3^2.5^8}\)

\(A=\dfrac{1.1.1.1}{4.1.1.1}\)

\(A=\dfrac{1}{4}\)

The number divisible by \(2\) is an even number.

=> The number divisible by \(7\) and \(2\) is an even number.

=> It is divisible by \(14\).

The least possible number is: \(14\).

The biggest possible number is: \(994\).

So there are: \(\left(994-14\right):14+1=71\left(numbers\right)\) satisfy this question.

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{97.98}-\dfrac{1}{98.99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{98.99}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{9702}\right)\)

\(=\dfrac{1}{2}.\dfrac{2425}{4851}\)

\(=\dfrac{2425}{9702}\)