I am testing the system.

[Test article] 21.1.2021

Answer as many questions as you can!
 

[test article]

In this page, Vietnamese is not allowed, post an English question and we'll help.

There are: \(3\times2\times5=30\)combinations

We have: \(A=87235-80106+\left(987+12\right)=7129+999=8128\)

But 8128 is not a square number so I think you should check your question again.

We have: \(\left(x\times x\right)^x\times x^x=\left(x\times x\times x\right)^x=\left(x\times x\times x\right)^x\) with \(x\in R\)

So..........

Sherlockkichi I have evidence to you. I know you have at least 5 accounts. I'll report this to Admin if you don't stop this. I'll post the evidence if anyone report this to me more. Thank you.

Thank you Huy Toàn 8A (TL)  and Lê Anh Duy have report this for me. I'll look further into this. I've searched Nobita - Kun and have found all 45-50 nicks from him. Admin will punish for anyone break the rules.

We have: \(x^2-3x+2=x^2-3x+2.25-0.25=x^2-2\times x\times1.5+\left(1.5\right)^2-0.25=\left(x-1.5\right)^2-0.25=0\)

\(\Rightarrow\left(x-1.5\right)^2=0.25\)

\(\Rightarrow x-1.5=\pm0.5\)

\(\Rightarrow x=\left\{1;2\right\}\)

So \(x=\left\{1;2\right\}\)

\(2x^3+6x^2=x^2+3x\)

\(2x\left(x^2+3x\right)=x^2+3x\)

Case 1: \(2x\left(x^2+3x\right)=0\) 

\(\Rightarrow x=0\)

Case 2: \(2x\left(x^2+3x\right)\ne0\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\dfrac{1}{2}\)

So \(x=\left\{0;\dfrac{1}{2}\right\}\)

Because Dao Trong Luan has posted the notification, I won't bother to keep this only for me.

Nobita - Kun (ducanh2007) has claimed to reach 320 faults in only 3 days. As to this page rules, this account will be deleted forever. I hope Admin will have a solution for this. 

Also, jimin bts cute (jimin recite) also has approx. 70 faults. I also hope Admin will have a way to improve this. Thank you.

For the smallest possible value of AE, A or E must be in the middle of the line. Consider A (E if you want) is that point. With AB = 3 and BC = 6, with A at the middle, then AC = 3. With CD = 8, AC = 3 and A at the middle, then AD = 5. With DE = 4, AD = 5 so D is in the middle, then AE = 1 (the smallest possible value of AE)

Every number has it's sum digits 10, so the odd one out is 1472.

The 1st shape has 3 lines. The 2nd shape has 4 lines, and so on. So the 8th shape has 10 lines. Because each line is 8cm, the perimeter of the 8th shape is: \(8\times10=80\left(cm\right)\)

Larry said:

a. “My BINGO has all even numbers.”

b. “My BINGO has two numbers evenly divisible by 5.”

c. “All of the numbers in my BINGO are evenly divisible by 4.” 

Lê Anh Duy Oh I forgot :)

We have: \(\dfrac{\left(4\times7+2\right)\left(6\times9+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)...\left(99\times102+2\right)}=\dfrac{5\times6\times7\times8\times...\times101\times102}{6\times7\times8\times9\times...\times100\times101}=5\times102=510\)