(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