\(1+2+3+....+x=500500\)
\(\Rightarrow\frac{x.\left(x+1\right)}{2}=500500\)
\(\Rightarrow\left(x+1\right).x=1001000=1000.1001\)
\(\Leftrightarrow\left(x+1\right).x=\left(1000+1\right).1000\)
\(\Leftrightarrow x=1000\)
Vậy \(x=1000\)
\(1+2+3+...+x=500500\)
\(\Rightarrow\frac{x\left(x+1\right)}{2}=500500\)
\(\Rightarrow x\left(x+1\right)=1001000\)
\(\Rightarrow x\left(x+1\right)=1000.1001\)
\(\Rightarrow x=1000\)
Vậy \(x=1000\)
Tìm số tự nhiên biết
1+2+3+...+x=500500
Đặt A=1+2+3+...+x
A số số hạng (x-1)\(\div\)1+1 =x(số hạng)
A=(x+1).x\(\div\)2
\(\Rightarrow\)(x+1).x\(\div\)2=500500
\(\Rightarrow\)(x+1).x=500500.2
=1001000
\(\Rightarrow\)(x+1).x=100100.10
\(\Rightarrow\) x=10
1+2+3+...+x=500500
Đặt A 1+2+3+...+x
A= (x+1).x\(\div\)2
\(\Rightarrow\)(x+1).x\(\div\)2=500500
\(\Rightarrow\)(x+1).x=500500.2
x=1001000
\(\Rightarrow\)(x+1).x=1001.1000
\(\Rightarrow\) x=1000