(x+1)+(x+2)+...+(x+9)+(x+10)=5
x + 1 + x + 2 + ... + x + 10 = 5
( x + x + ...+ x ) + ( 1 + 2 + ... + 10 ) = 5
Số số hạng là : ( 10 - 1 ) : 1 + 1 = 10 ( số )
Tổng là : ( 10 + 1 ) . 10 : 2 = 55
Ta có :
10x + 55 = 5
10x = -50
x = -50 : 10
x = -5
Vậy,...........
(x+1) + (x+2) + ...+ (x+9)+(x+10) = 5
x.10 + (1+2+...+9+10) = 5
x.10 + 55 =5
x.10 = 50
x = 5
k mk nha
(x + 1) + (x + 2) +...+ (x + 9) + (x + 10) = 5
=> x + 1 + x + 2 + ... + x + 9 + x + 10 = 5
=> (x + x + ... + x + x) + (1 + 2 + ... + 9 + 10) = 5
=> 10x + 55 = 5
=> 10x = -50
=> x = -5
vậy_
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=5\)
\(\Rightarrow\left(x+x+...+x\right)+\left(1+2+3+...+9+10\right)=5\)
\(\Rightarrow10x+55=5\)
\(\Rightarrow10x=-50\)
\(\Rightarrow x=-5\)
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+9\right)+\left(x+10\right)=5\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+..+9+10\right)\)
\(\Leftrightarrow10x+55=5\)
\(\Leftrightarrow10x=5-55\)
\(\Leftrightarrow10x=-50\)
\(\Leftrightarrow x=\frac{-50}{10}=-5\)
\(\left(x+1\right)+\left(x+2\right)+....+\left(x+9\right)+\left(x+10\right)=5\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+...+9+10\right)=5\)
\(\Leftrightarrow10x+\frac{\left[\left(10-1\right):1+1\right]\cdot\left(10+1\right)}{2}=5\)
\(\Leftrightarrow10x+55=5\)
\(\Leftrightarrow10x=-50\)
\(\Leftrightarrow x=-5\)
Vậy x=-5