Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1+2+3+4+...+x=500500\Rightarrow\frac{x\cdot\left(x+1\right)}{2}=500500\Rightarrow x\cdot\left(x+1\right)=1001000=1000.1001\)
ta có:
1 + 2 + 3 +...+ x = 500500
=> (x + 1)x : 2 = 500500
=> (x + 1)x = 500500.2 = 1001000
=> (x + 1)x = 1000.1001
=> x = 1000
\(1+2+3+....+x=500500\)
\(\Leftrightarrow\left(1+x\right).x\div2=500500\)
\(\Leftrightarrow\left(1+x\right).x=500500.2\)
\(\Leftrightarrow\left(1+x\right).x=1001000\)
\(\Leftrightarrow\left(1+x\right).x=1001.1000\)
\(\Leftrightarrow x=1000\)
Vậy x = 1000
x>0
ta có 1+2+3+...+x= (x+1).x /2
Nên (x+1)x/2=500500
(x+1)x =500500.2=1001000
x2+x - 1001000=0
\(\orbr{\begin{cases}x=1000\\x=-1001\end{cases}}\)
Vậy x=1000
\(1+2+3+...+x=500500\)
\(\Rightarrow\frac{x.\left(x+1\right)}{2}=500500\)
\(\Rightarrow x.\left(x+1\right)=1001000\)
\(\Rightarrow1000.1001\)
..
1+2+3+4+...+x=500500
\(\frac{\left(x+1\right).x}{2}=500500\)
\(\left(x+1\right).x=1001000\)
\(1001.1000=100100\)
Vậy x = 1000
1+2+3+4+...+X = 500500
( X + 1 ) x X : 2 = 500500
( X + 1) x X = 500500 x 2
( X + 1) x X = 1001000
1000 x ( 1000+ 1) = 1001000
Vậy x = 1000
\(1+2+3+....+x=500500\)
\(\Rightarrow\frac{x.\left(x+1\right)}{2}=500500\)
\(\Rightarrow\left(x+1\right).x=1001000=1000.1001\)
\(\Leftrightarrow\left(x+1\right).x=\left(1000+1\right).1000\)
\(\Leftrightarrow x=1000\)
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 \(x=1000\)
a, (x+1)+(x+2)+(x+3)+...+(x+100) = 7450
(x+x+...+x)+(1+2+...+100) = 7450
100 x + 101 . 100 2 = 7450
100x = 2400
x = 24
b, 1+2+3+...+x = 500500
Đặt: A = 1+2+3+...+x
số hạng A (x - 1) : 1 + 1 = x
Tổng của A
A = x + 1 . x 2 = 500500
(x+1).x = 1001000
Ta thấy
1000.1001 = 1001000
=> x = 1000
=>\(\frac{n.\left(n-1\right)}{2}=500500\)
n(n-1)=500500.2
=>n(n-1)=1001000
=>n(n-1)=1001.(1001-1)
=>n=1001