Ta có: 5 = 2+3; 9 = 4+5 ; 13 = 6+7 ; 17 = 8+9;.....

Do vậy x = a + ( a + 1) ( a thuộc N )

Nên 1 + 5 + 9 + 13 + 16 + ....+ x = 1+2+3+4+5+6+7+.....+a+ ( a + 1 ) = 501501

Hay (a + 1)( a + 2) = 1003002 = 1001 . 1002

Suy ra : a = 1000

Do đó : x = 1000 + ( 1000+ 1) = 2001

Vậy, x= 2001

vi sao ma (a+1)(a+2) = 1003002

Ta có: \(\left(x+1\right).x\div2=501501\)

          \(\left(x+1\right).x=501501.2\)

          \(\left(x+1\right).x=10003002\)

          \(\left(x+1\right).x=10003002=\)

Ta có: 5 = 2+3; 9 = 4+5 ; 13 = 6+7 ; 17 = 8+9;.....

Do vậy x = a + ( a + 1)  ( a thuộc N )

Nên 1 + 5 + 9 + 13 + 16 + ....+ x = 1+2+3+4+5+6+7+.....+a+ ( a + 1 ) = 501501

Hay (a + 1)( a + 2)  = 1003002 = 1001 . 1002 

Suy ra : a = 1000

Do đó : x = 1000 + ( 1000+ 1) = 2001

tao ko biet

x=2001

Ta có: 5 = 2+3; 9 = 4+5 ; 13 = 6+7 ; 17 = 8+9;.....

Do vậy x = a + ( a + 1) ( a thuộc N )

Nên 1 + 5 + 9 + 13 + 16 + ....+ x = 1+2+3+4+5+6+7+.....+a+ ( a + 1 ) = 501501

Hay (a + 1)( a + 2) = 1003002 = 1001 . 1002

Suy ra : a = 1000

Do đó : x = 1000 + ( 1000+ 1) = 2001 

Vậy......

mik ko pk

\(1+5+9+13+17+...+x=501501\)

\(1+2+3+4+5+6+7+...+a+\left(a+1\right)=501501\)

\(\left(a+1\right)\left(a+1+1\right):2=501501\)

\(\left(a+1\right)\left(a+2\right):2=501501\)

\(\left(a+1\right)\left(a+2\right)=1003002\)

\(\left(a+1\right)\left(a+2\right)=1001\cdot1002\)

\(\Rightarrow a=1000\)

Vậy \(x=a+a+1=1000+1000+1=2001\)

~Học tốt~

bủhh

Ta có 5= 2+3; 9=4+5; 13=6+7; 17= 8+9;...

Do vậy x= a+ (a+1) ( a thuộc N)

=> 1+5+9+13+17+....+ x= 1+2+3+4+5+6+7+8+9+.... + a+ (a+1)= 501501

Hay (a+1) (a+2) = 1003002= 1001.1002

=> a=1000

Vậy x =1000+ (1000+1) =2001

Tick cho mình nha