Question

Tìm x \(\in\)N

a) x + ( x + 1 ) + ( x + 2 ) + .....+ ( x + 30 ) = 1240

b) 1 + 2 + 3 +........+ x = 210

Answer 1

a) x + ( x + 1 ) + .....+ ( x + 30 ) = 1240

=> ( x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240

=> 31 x + 465 = 1240

=> 31 x = 775

=> x = 25.

b) 1 + 2 + 3 + ... + x = 210

=> ( x + 1 ) . [ ( x - 1 ) : 1 + 1 ] : 2 = 210

=> ( x + 1 ) . x = 420

=> ( x + 1 ) . x = 21 . 20

=> x = 20

Answer 2

a) x+(x+1)+...+(x+30)=1240

(x+...+x)+(1+2+...+30)=1240

31x+465=1240

31x=775

x=25

b) 1+2+3+...+x=210

(x+1).[(x-1):1+1]:2=210

(x+1).x=420

(x+1).x=21.20

x = 20.