(x+1)+(x+2)+(x+3)+...+(x+30)=795
<=> (x+x+x+...+x)+(1+2+3+...+30)=795
<=> 30x+[(30+1) x 30:2]=795
<=> 30x+465=795
<=> 30x=330
<=> x=11
Ta có : ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 795
<=> ( x + x + x +... + x ) + ( 1 + 2 + ... + 30 ) = 795
<=> 30.x + 465 = 795
<=> 30 .x = 795- 465
<=> 30 . x = 330
<=> x = 330 : 30
<=> x = 11
Vậy x = 11
\(\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=795\)
\(=>x.30+\left(1+2+3+...+30\right)=795\)
\(=>30x+\frac{\left(30+1\right).30}{2}=795\)
\(=>30x+15.30=795\)
\(=>30x+450=795\)
\(=>x=\frac{795-450}{30}=\frac{345}{30}=\frac{23}{2}\)