1 + 2 + 3 + ... + x = 465 (x > 0)
x.(x + 1) : 2 = 465
x.(x + 1) = 465 . 2
x² + x = 930
x² + x - 930 = 0
x² + 31x - 30x - 930 = 0
(x² + 31x) - (30x + 930) = 0
x(x + 31) - 30(x + 31) = 0
(x + 31)(x - 30) = 0
x + 31 = 0 hoặc x - 30 = 0
*) x + 31 = 0
x = -31 (loại)
*) x - 30 = 0
x = 30 (nhận)
Vậy x = 30
\(1+2+3+...+x=465\) (ĐK: \(x\in N\))
\(\Rightarrow\left(x+1\right)\left[\left(x-1\right):1+1\right]:2=465\)
\(\Rightarrow\left(x+1\right)\left(x-1+1\right):2=465\)
\(\Rightarrow\left(x+1\right)\cdot x:2=465\)
\(\Rightarrow x\left(x+1\right)=465\cdot2\)
\(\Rightarrow x\cdot\left(x+1\right)=930\)
Mà: \(30\cdot31=930\)
Nên: \(x=30\)
Vậy: x=30
