Question

Tìm x biết ( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 28 ) = 155

Tớ nghĩ mãi ko ra chỉ jup với 

Answer 1

=> ( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 28 ) = 155

=> (x + x + x + ... + x) + (1 + 2 + 3 + ... + 28) = 155

=> 28x + \(\frac{28.29}{2}\) = 155

=> 28x + 406 = 155

=> 28x = 155 - 406 = -251

=> x = \(-\frac{251}{28}\)

Answer 2

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+28\right)=155\)

\(x+1+x+2+x+3+...+x+28=155\)

\(28x+\left(1+2+3+......+28\right)\)

\(28x+406=155\)

\(28x=-251\)

\(x=-\frac{251}{28}\)

Answer 3

( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 28 ) = 155

(x + x + x + ... + x + x) + (1 + 2 + 3 + ... + 27 + 28) = 155

28x + 406 = 155

28x = 155 - 406

28x = -251

x = -251/28

Answer 4

\((x+1)+(x+2)+(x+3)+......+(x+28)=155\)

\(x+1+x+2+x+3+.........+x+28=155\)

\((x+x+x+x+.........+x)+(1+2+3+........+28)=155\)

\(28x+1+2+3+4+.......+28=155\)

\(28x+(28+1).28:2=155\)

\(28x+406=155\)

\(28x=155-406\)

\(28x=-251\)

\(x=-251:28\)

\(x=\dfrac{-251}{28}\)