(x+1)+(x+2)+...+(x+30)=795
(x+x+x+...+x) + (1+2+3+...+30) = 795
30x + 465 = 795
30x = 795 - 465
30x = 330
x = 330 : 30
x = 11
Vậy x = 11
tho T-T
\((x+1)+(x+2)+...+(x+30)=795\\\Rightarrow (x+x+...+x)+(1+2+...+30)=795 (1)\)
Đặt \(A=1+2+...+30\)
Số các số hạng trong tổng \(A\) là:
\(\left(30-1\right):1+1=30\left(số\right)\)
Tổng \(A\) bằng:
\(\left(30+1\right)\cdot30:2=465\)
Thay \(A=465\) vào \(\left(1\right)\), ta được:
\(30x+465=795\\\Rightarrow 30x=795-465\\\Rightarrow 30x=330\\\Rightarrow x=330:30\\\Rightarrow x=11\)
#\(Toru\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=795\)
\(\Leftrightarrow30x+\left(1+2+...+30\right)=795\)
\(\Leftrightarrow30x+\dfrac{30\cdot31}{2}=795\)
\(\Leftrightarrow30x=795-\dfrac{30\cdot31}{2}=330\)
\(\Leftrightarrow x=\dfrac{330}{30}=11.\)