khoan cach giua 2 so la :2-1=1
co tat ca so so hang hay x la :(10-1):1+1=10
(x+1)+(x+2)+.....................(x+10)=2015
x*10+[(1+10)*10/2]=2015
x*10+55 =2015
x*10 =2015-55
x*10 =1960
x =1960/10
x =196
tìm x biết:( x+1) + ( x+ 2) + ( x+3)+ ( x+4)+.....+( x+9)+ ( x+10) = 2015
Giải:Ta có:\(\left(x+1\right)+\left(x+2\right)+......+\left(x+10\right)=2015\)
\(\Rightarrow x+1+x+2+.....+x+10=2015\)
\(\Rightarrow10x+\left(1+2+......+10\right)=2015\)
\(\Rightarrow10x+55=2015\Rightarrow10x=2015-55\)
\(\Rightarrow10x=1960\Rightarrow x=196\)
Vậy .....................
(x+1) + ( x+2 ) + ( x+3) + ( x+4) + ...........+ (x+9) + (x+ 10) =2015
x* 10 + ( 1+ 2 +3 +4 + ........+9 +10) = 2015
x * 10 + 55 = 2015
x * 10 = 2015-55
x* 10 =1960
x = 1960 /10
x = 196
Ta có:
( x + x + x + x + ....... + x + x ) + ( 1 + 2 + 3 + 4 + .... + 9 + 10 )
( x + x + x + x + ....... + x + x ) + [ ( 10 - 1 ) : 1 + 1 ]
( x + x + x + x + ....... + x + x ) + ( 10 + 1 ) x 10 :2
( x + x + x + x + ....... + x + x ) + 55
Mỗi số hạng chứa 1 x. Ta có: 10 số hạng là có 10 x
X là : 55 : 10 = 5,5
=> (x + x + ... + x) + (1 + 2 + ... + 10) =2015
10 số x
=> 10x + 55 = 2015
=> 10x = 1960
=> x = 196
Vậy số đó là 196
(x+x+x+... +x) + (1+2+3+... +9+10) = 2015
10x + 55 = 2015
10x = 2015 - 55
10x = 1960
x = 1960 : 10
x = 196
