2 người đó là 1 người.Và theo tui đc biết thì chủ nhân nick đó thường lấy 2 nick tick đi tick lại ,r nhờ ai đó(hoặc đã đc họ tick tự nguyện) nên đc GP

_____________Linh__________________

Đặt:

\(\dfrac{x}{2}=\dfrac{y}{5}=t\Rightarrow\left\{{}\begin{matrix}x=2t\\y=5t\end{matrix}\right.\)

Khi đó:

\(xy=100\Leftrightarrow2t.5t=100\)

Suy ra : \(10t^2=100\Leftrightarrow t^2=10\) suy ra \(\left[{}\begin{matrix}t=\sqrt{10}\\t=-\sqrt{10}\end{matrix}\right.\)

\(\circledast\) nếu \(t=\sqrt{10}\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{10}.2=\sqrt{40}\\y=\sqrt{10}.5=\sqrt{250}\end{matrix}\right.\)

\(\circledast\) nếu \(t=-\sqrt{10}\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{10}.2=-\sqrt{40}\\y=-\sqrt{10}.5=-\sqrt{250}\end{matrix}\right.\)

x :1/2=2/3*3/4

x :1/2=1/2

x       =1/2*1/2

x       =1/4

tick di

x : 1/2= 2/3*3/4

x : 1/2= 1/2

x = 1/2 * 1/2

x = 1/4

\(A=\left(2n-1\right)^3-2n+1\)

\(A=8n^3-6n+6n-1-2n+1\)

\(A=8n^3-2n=2n\left(4n^2-1\right)\)

\(A=2n\left(2n+1\right)\left(2n-1\right)\)

\(A=\left(2n-1\right)2n\left(2n+1\right)⋮6\) ( 3 số tự nhiên liên tiếp)

\(100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+....+\left(2-1\right)\left(2+1\right)\)

\(=1.199+1.195+...+1.3\)

\(=199+195+....+3\)

\(=\left[\left(\dfrac{199-3}{4}\right)+1\right]:2.\left(199+3\right)=5050\)

\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{3^{32}-1}{2}\)

\(3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{64}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)\)

\(=\left(2^8-1\right)......\left(2^{64}+1\right)=2^{128}-1\)

I like reading conan.Because it very interesting and character is very cute.